Intersection of a Circle and a Rectangular Hyperbola
If a circle x 2 + y2 + 2gx + 2 fy + k = 0 cuts a rectangular hyperbola xy = c 2 in A, B, C and D and the
parameters of these four points be t1, t2, t3 and t4 respectively; then
(1) (i) ∑ t1 = − 2g (ii) ∑ t1t2 = k
c c2
(iii) ∑ t1t2t3 = − 2f (iv) t1t 2t3 t4 = 1 (v) 1 2f
c ∑ t1 = − c
(2) Orthocentre of ∆ABC is H − ct 4 , −c but D is ct 4 , c
t4 t4
Hence H and D are the extremities of a diagonal of rectangular hyperbola.
(3) Centre of mean position of four points is c ∑ t1, c ∑ 1 i.e., − g , − f
4 4 t1 2 2
Centres of the circles and rectangular hyperbola are (– g, – f) and (0, 0); mid point of centres of circle and
hyperbola is − g , − f . Hence the centre of the mean position of the four points bisects the distance between the
2 2
centres of the two curves (circle and rectangular hyperbola)
(4) If the circle passing through ABC meet the hyperbola in fourth points D; then centre of circle is (–g, –f)
i.e., c t1 + t2 + t3 + 1 ; c 1 + 1 + 1 + t1 t 2 t 3
2 t1t 2 t 3 2 t1 t2 t3
Example: 39 If a circle cuts a rectangular hyperbola xy = c2 in A, B, C, D and the parameters of these four points be t1, t2, t3 and t4
Solution: (b)
respectively. Then [Kurukshetra CEE 1998]
(a) t1t2 = t3t4 (b) t1t2t3t4 = 1 (c) t1 = t2 (d) t3 = t4
Let the equation of circle be x2 + y2 = a2 ......(i)
Parametric equation of rectangular hyperbola is x = c t, y = c
t
Put the values of x and y in equation (i) we get c2t2 + c2 = a2 ⇒ c2t4 − a2t2 + c2 =0
t2
Hence product of roots t1t2t3t4 = c2 =1
c2
Example: 40 If the circle x2 + y2 = a2 intersects the hyperbola xy = c2 in four points P (x1, y1), Q(x2, y2), R(x3, y3) , S(x4, y4 ) then
[IIT 1998]
(a) x1 + x2 + x3 + x4 = 0 (b) y1 + y2 + y3 + y4 = 0 (c) x1x2x3x4 = c4 (d) y1y2y3y4 = c4
Solution: (a,b,c,d) Given, circle is x2 + y2 = a2 .......(i) and hyperbola be xy = c2 .....(ii)
from (ii) y= c2 . Putting in (i), we get x2 + c4 = a2 ⇒ x4 − a2x2 + c4 =0
x x2
∴ x1 + x2 + x3 + x4 = 0 , x1x2x3x4 = c4
Since both the curves are symmetric in x and y, ∴ y1 + y2 + y3 + y4 = 0 ; y1y2y3y4 = c4 .
Co-ordinate geometry of three dimensions
Example: 31 The length of the perpendicular from the origin to line r = (4i + 2j + 4k) + λ(3i + 4j − 5k) is [AMU 1992]
Solution: (d) B
(a) 2 5 (b) 2 (c) 5 2 (d) 6
α = 0.i + 0.j + 0.k P( α→)
L
PL = (a − α) − (a − α).b b
| b |2
A
r=a+λb
PL = (4i + 2j + 4k) − (4 i + 2j + 4k).(3i + 4j − 5k) (3i + 4j − 5k) = 4i + 2j + 4k − 12 +8− 20 .(3i + 4j − 5k)
9 + 16 + 25 50
PL = 4i + 2j + 4k
The length of PL is magnitude of PL i.e., Length of perpendicular =| PL|= 16 + 4 + 16 = 6 .
Example: 32 The image of point (1, 2, 3) in the line r = (6i + 7j + 7k) + λ(3i + 2j − 2k) is
Solution: (d)
(a) (5, –8, 15) (b) (5, 8, –15) (c) (–5, –8, –15) (d) (5, 8, 15)
Given that, a = 6i + 7j + 7k , b = 3i + 2j − 2k and α = i + 2j + 3k P(→α
Then, β= 2a − 2(a − α).b b −α
| b|2
A L B
(a+λb)
2(5i 5j + 4k).(3i + 2j 2k) (3i
= 2(6i + 7j + 7k) − + 9+4+4 − + 2j − 2k) − (i + 2j + 3k) Q(→β )
On solving, β = 5i + 8j + 15k . Thus β is the position vector of Q, which is the image of P in given line.
Hence image of point (1, 2, 3) in the given line is (5, 8, 15).
Shortest distance between two straight lines.
(1) Skew lines : Two straight lines in space which are neither parallel nor intersecting are called skew lines.
Thus, the skew lines are those lines which do not lie in the same plane. Q
l2 Line of shortest
distance
l1 P
(2) Line of shortest distance : If l1 and l2 are two skew lines, then the straight line which is perpendicular
to each of these two non-intersecting lines is called the “line of shortest distance.”
Note : There is one and only one line perpendicular to each of lines l1 and l2 .
(3) Shortest distance between two skew lines
(i) Cartesian form : Let two skew lines be x − x1 = y − y1 = z − z1 and x − x2 = y − y2 = z − z2
l1 m1 n1 l2 m2 n2
Therefore, the shortest distance between the lines is given by
x2 − x1 y2 − y1 z2 − z1
l1 m1 n1
d = l2 m2 n2
(m1n2 − m2n1)2 + (n1l2 − l1n2)2 + (l1m2 − l2m1)2
(ii) Vector form : Let l1 and l 2 be two lines whose equations are l1 : r = a1 + λb1 and l2 : r = a2 + µb2
respectively. Then, Shortest distance PQ = (b1 × b 2 ).(a 2 − a1 ) = |[b 1 b2 (a 2 − a1 )]
|b1 × b2 | |b1 ×b 2|
(4) Shortest distance between two parallel lines : The shortest distance between the parallel lines
r = a1 + λb and r = a2 + µb is given by d = |(a 2 − a 1) × b | .
| b |
(5) Condition for two lines to be intersecting i.e. coplanar
(i) Cartesian form : If the lines x − x1 = y − y1 = z − z1 and x − x2 = y − y2 = z − z2 intersect, then
l1 m1 n1 l2 m2 n2
x2 − x1 y2 − y1 z2 − z1
l1 m1 n1 = 0 .
l2 m2 n2
(ii) Vector form : If the lines r = a1 + λb1 and r = a 2 + λb 2 intersect, then the shortest distance between
them is zero. Therefore, [b1b 2 (a 2 − a1)] = 0 ⇒ [(a2 − a1) b1b2] = 0 ⇒ (a 2 − a1).(b1 × b 2 ) = 0
Important Tips
• Skew lines are non-coplanar lines.
• Parallel lines are not skew lines.
• If two lines intersect, the shortest distance (SD) between them is zero.
• Length of shortest distance between two lines is always taken to be positive.
• Shortest distance between two skew lines is perpendicular to both the lines.
(6) To determine the equation of line of shortest distance : To find the equation of line of shortest
distance, we use the following procedure :
(i) From the given equations of the straight lines,
i.e. x − a1 = y − b1 = z − c1 =λ (say) ……(i)
l1 m1 n1 ……(ii)
and x − a2 = y − b2 = z − c2 =µ (say)
l2 m2 n2
Find the co-ordinates of general points on straight lines (i) and (ii) as
(a1 + λl1,b1 + λm1, c1 + λn1 ) and (a2 + µl2 ,b2 + µm2 , c 2 + µn2 ) .
(ii) Let these be the co-ordinates of P and Q, the two extremities of the length of shortest distance. Hence, find
the direction ratios of PQ as (a2 + l2µ) − (a1 + l1λ), (b2 + m2µ) − (b1 + m1λ), (c2 + m2µ) − (c1 + n1λ) .
(iii) Apply the condition of PQ being perpendicular to straight lines (i) and (ii) in succession and get two
equations connecting λ and µ. Solve these equations to get the values of λ and µ.
(iv) Put these values of λ and µ in the co-ordinates of P and Q to determine points P and Q.
(v) Find out the equation of the line passing through P and Q, which will be the line of shortest distance.
Note : The same algorithm may be observed to find out the position vector of P and Q, the two
extremities of the shortest distance, in case of vector equations of straight lines. Hence, the line of
shortest distance, which passes through P and Q, can be obtained.
Example: 33 The shortest distance between the lines x −1 = y−2 = z−3 and x− 2 = y −4 = z−5 is [Kerala (Engg.)2001; DCE 1993]
2 3 4 3 4 5
(a) 1 (b) 1 (c) 1 (d) 1
6 6 3 3
2−1 4 −2 5−3 122
234 234
Solution: (b) S.D. = 34 5 3 4 5 1.
= =
(15 − 16)2 + (12 − 10)2 + (8 − 9)2 1 + 1 + 4 6
Example: 34 The shortest distance between the lines r = (i + j − k) + λ(3i − j) and r = (4i − k) + µ(2i + 3k) is [Pb. CET 1995]
(a) 6 (b) 0 (c) 2 (d) 4
Solution: (b) S.D. = (b1 × b2).(a2 − a1) = [(3i − j)× (2i + 3k)].(3i − j) = (−3i − 9j + 2k).(3i − j) = −9+9+0 .
Example: 35 | b1 × b2 | |(3i − j)× (2i + 3k)| 9 + 81 + 4 94
Solution: (c)
Hence, S.D. = 0
The line x−2 = y−3 = z−4 and x −1 = y−4 = z −5 are coplanar, if [AIEEE 2003]
1 1 −k k 2 1
(a) k = 0 or –1 (b) k = 0 or 1 (c) k = 0 or –3 (d) k = 3 or –3
Lines are coplanar, if
x2 − x1 y2 − y1 z2 − z1 =0 ⇒ 1− 2 4−3 5−4
l1 m1 n1 1 1 − k = 0 ⇒ k2 + 3k = 0 ⇒ k(k + 3) = 0 ⇒ k = 0 , k = −3
l2 m2 n2 k 2
1
Example: 36 The lines r = a + λ(b × c) and r = b + µ(c × a) will intersect if
(a) a × c = b × c (b) a.c = b.c (c) b × a = c × a (d) None of these
Solution: (b) If lines are intersecting, then
(a 2 − a1).(b1 × b 2 ) = 0 ⇒ b (a − b).[(b × c)× (c × a)] = 0
⇒ (a − b).[(b × c.a)c − (b × c.c)a] = 0 ⇒ (a − b)[(b × c.a)c] = 0
⇒ [(a − b).c]abc = 0 ⇒ (a.c − b.c)(abc) = 0 ⇒ a.c − bc = 0 ⇒ a.c = b.c
Example: 37 If the straight lines x = 1 + s, y = 3 − λs, z = 1 + λs and x= t , y = 1+ t, z = 2 − t , with parameters s and t respectively, are
2
co-planar, then λ equals [AIEEE 2004]
(a) 0 (b) –1 (c) 1 (d) –2
−2
Solution: (d) We have x −1 = y+3 = z −1 = s and 2x = y −1 = z−2 = t
1 −λ λ 1 1 −1
i.e. x − 0 = y − 1 = z−2 = t
1 2 −2 2
Since, lines are co-planar,
Then, x2 − x1 y2 − y1 z2 − z1 =0 ⇒ −1 4 1
l1 m1 n1 1 −λ λ =0
l2 m2 n2 12 −2
On solving, λ = −2 .
The Plane
Definition of plane and its equations.
If point P(x, y, z) moves according to certain rule, then it may lie in a 3-D region on a surface or on a line or it
may simply be a point. Whatever we get, as the region of P after applying the rule, is called locus of P. Let us
discuss about the plane or curved surface. If Q be any other point on it’s locus and all points of the straight line PQ
lie on it, it is a plane. In other words if the straight line PQ, however small and in whatever direction it may be, lies
completely on the locus, it is a plane, otherwise any curved surface.
(1) General equation of plane : Every equation of first degree of the form Ax + By + Cz + D = 0 represents
the equation of a plane. The coefficients of x, y and z i.e. A, B, C are the direction ratios of the normal to the plane.
(2) Equation of co-ordinate planes Y
XOY-plane : z = 0 XOY-plane
YOZ -plane : x = 0 YOZ-plane
ZOX-plane : y = 0
X
ZOX-plane
(3) Vector equation of plane Z
(i) Vector equation of a plane through the point A(a) and perpendicular to the vector n is (r − a).n = 0 or
r.n = a.n A(a)
N
Note : The above equation can also be written as r.n = d , where P(r) a
n
d = a.n . This is known as the scalar product form of a plane. r
(4) Normal form : Vector equation of a plane normal to unit vector nˆ N P(r)
and at a distance d from the origin is r.nˆ = d .
Note : If n is not a unit vector, then to reduce the equation r.n = d
to normal form we divide both sides by |n| to obtain r ⋅ | n | = | d | or dn r
n n O
r.nˆ = | d | .
n
(5) Equation of a plane passing through a given point and parallel to two given vectors : The
equation of the plane passing through a point having position vector a and C
c
parallel to b and c is r = a + λb + µc , where λ and µ are scalars.
M P(r)
(6) Equation of plane in various forms P(a) L b B
(i) Intercept form : If the plane cuts the intercepts of length a, b, c on co-
ordinate axes, then its equation is x + y + z = 1.
a b c
(ii) Normal form : Normal form of the equation of plane is lx + my + nz = p ,
where l, m, n are the d.c.’s of the normal to the plane and p is the length of perpendicular from the origin.
(7) Equation of plane in particular cases
(i) Equation of plane through the origin is given by Ax + By + Cz = 0 .
i.e. if D = 0, then the plane passes through the origin.
(8) Equation of plane parallel to co-ordinate planes or perpendicular to co-ordinate axes
(i) Equation of plane parallel to YOZ-plane (or perpendicular to x-axis) and at a distance ‘a’ from it is x = a.
(ii) Equation of plane parallel to ZOX-plane (or perpendicular to y-axis) and at a distance ‘b’ from it is y = b.
(iii) Equation of plane parallel to XOY-plane (or perpendicular to z-axis) and at a distance ‘c’ from it is z = c.
Important Tips
• Any plane perpendicular to co-ordinate axis is evidently parallel to co-ordinate plane and vice versa.
• A unit vector perpendicular to the plane containing three points A, B, C is AB × AC .
| AB × AC|
(9) Equation of plane perpendicular to co-ordinate planes or parallel to co-ordinate axes
(i) Equation of plane perpendicular to YOZ-plane or parallel to x-axis is By + Cz + D = 0 .
(ii) Equation of plane perpendicular to ZOX-plane or parallel to y axis is Ax + Cz + D = 0 .
(iii) Equation of plane perpendicular to XOY-plane or parallel to z-axis is Ax + By + D = 0 .
(10) Equation of plane passing through the intersection of two planes
(i) Cartesian form : Equation of plane through the intersection of two planes
P = a1 x + b1y + c1z + d1 = 0 and Q = a2 x + b2y + c 2z + d2 = 0 is P + λQ = 0 , where λ is the parameter.
(ii) Vector form : The equation of any plane through the intersection of planes r.n1 = d1 and r.n 2 = d2 is
r.(n1 + λn 2 ) = d1 + λd2 , where λ is an arbitrary constant.
(11) Equation of plane parallel to a given plane
(i) Cartesian form : Plane parallel to a given plane ax + by + cz + d = 0 is ax + by + cz + d′ = 0 , i.e. only
constant term is changed.
(ii) Vector form : Since parallel planes have the common normal, therefore equation of plane parallel to
plane r.n = d1 is r.n = d2 , where d2 is a constant determined by the given condition.
Equation of plane passing through the given point.
(1) Equation of plane passing through a given point : Equation of plane passing through the point
(x1, y1, z1) is A(x − x1) + B(y − y1) + C(z − z1) = 0 , where A, B and C are d.r.’s of normal to the plane.
(2) Equation of plane through three points : The equation of plane passing through three non-collinear
x y z 1 x − x1 y − y1 z − z1
x2 − x1 y2 − y1 z2 − z1 = 0 .
points (x1, y1, z1) , (x2, y2, z2 ) and (x3 , y3 , z3 ) is x1 y1 z1 1 =0 or x3 − x1 y3 − y1 z3 − z1
x2 y2 z2 1
x3 y3 z3
1
7.19 Foot of perpendicular from a point A(α, β, γ) to a given plane ax + by + cz + d = 0.
If AP be the perpendicular from A to the given plane, then it is parallel to the normal, so that its equation is
x −α = y−β = z −γ =r (say)
a b c
Any point P on it is (ar + α,br + β,cr + γ ). It lies on the given plane and we find the value of r and hence the
point P.
(1) Perpendicular distance
(i) Cartesian form : The length of the perpendicular from the point P(x1, y1, z1) to the plane
ax + by + cz + d = 0 is ax1 + by1 + cz1 + d .
a2 + b2 + c2
Note : The distance between two parallel planes is the algebraic difference of perpendicular distances on the
planes from origin.
Distance between two parallel planes Ax + By + Cz + D1 = 0 and Ax + By + Cz + D2 = 0 is
D2 ~ D1 .
A2 + B2 + C2
(ii) Vector form : The perpendicular distance of a point having position vector a from the plane r.n = d is
given by p = | a.n − d |
|n|
(2) Position of two points w.r.t. a plane : Two points P(x1, y1, z1) and Q(x2, y2, z2 ) lie on the same or
opposite sides of a plane ax + by + cz + d = 0 according to ax1 + by1 + cz1 + d and ax2 + by2 + cz2 + d are of same
or opposite signs. The plane divides the line joining the points P and Q externally or internally according to P and Q
are lying on same or opposite sides of the plane.
Angle between two planes.
(1) Cartesian form : Angle between the planes is defined as angle between normals to the planes drawn
from any point. Angle between the planes a1x + b1y + c1z + d1 = 0 and a2 x + b2y + c2z + d2 = 0 is
cos −1 a1a2 + b1b2 + c1c2
(a12 + b12 + c12 )(a 2 + b22 + c 2 )
2 2
Note : If a1a2 + b1b2 + c1c2 = 0 , then the planes are perpendicular to each other.
If a1 = b1 = c1 , then the planes are parallel to each other.
a2 b2 c2
(2) Vector form : An angle θ between the planes r1 .n 1 = d1 and r2 .n 2 = d2 is given by cosθ = ±| n1.n 2 | .
n1 ||n
2
Equation of planes bisecting angle between two given planes .
(1) Cartesian form : Equations of planes bisecting angles between the planes a1 x + b1y + c1z + d1 = 0 and
a2x + b2y + c2z + d = 0 are a1 x + b1y + c1z + d1 = ± a2x + b2y + c2z + d2 .
(a12 + b12 + c12 ) (a 2 b 2 c 2 )
2 + 2 + 2
Note : If angle between bisector plane and one of the plane is less than 45o, then it is acute angle bisector,
otherwise it is obtuse angle bisector.
If a1a2 + b1b2 + c1c2 is negative, then origin lies in the acute angle between the given planes
provided d1 and d2 are of same sign and if a1a2 + b1b2 + c1c 2 is positive, then origin lies in the
obtuse angle between the given planes.
(2) Vector form : The equation of the planes bisecting the angles between the planes r1.n1 = d1 and
r2 .n 2 = d2 are | r.n1 − d1 | = | r.n 2 −d 2 | or r.n1 − d1 = ± r.n 2 − d2 or r.(nˆ 1 ± nˆ ) = d1 ± d2 .
|n1 | | 2| |n1 | |n2 | n1 n2
n 2 | | | |
Image of a point in a plane.
Let P and Q be two points and let π be a plane such that
(i) Line PQ is perpendicular to the plane π, and
(ii) Mid-point of PQ lies on the plane π.
Then either of the point is the image of the other in the plane π.
To find the image of a point in a given plane, we proceed as follows
(i) Write the equations of the line passing through P and normal to the given plane as
x − x1 = y − y1 = z − z1 . P(x1,y1,z1)
a b c ax+by+cz+d=0
(ii) Write the co-ordinates of image Q as (x1 + ar, y1, + br, z1 + cr) . R
(x1+ar,y1+br,z1+cr) Q
(iii) Find the co-ordinates of the mid-point R of PQ. π
(iv) Obtain the value of r by putting the co-ordinates of R in the equation
of the plane.
(v) Put the value of r in the co-ordinates of Q.
Coplanar lines.
Lines are said to be coplanar if they lie in the same plane or a plane can be made to pass through them.
(1) Condition for the lines to be coplanar
(i) Cartesian form : If the lines x − x1 = y − y1 = z − z1 and x − x2 = y − y2 = z − z2 are coplanar
l1 m1 n1 l2 m2 n2
Then x2 − x1 y2 − y1 z2 − z1 = 0.
l1 m1 n1
l2 m2 n2
The equation of the plane containing them is x − x1 y − y1 z − z1 = 0 or x − x2 y − y2 z − z2 = 0.
l1 m1 n1 l1 m1 n1
l2 m2 n2 l2 m2 n2
(ii) Vector form : If the lines r = a1 + λb1 and r = a 2 + λb 2 are coplanar, then [a1b1b 2 ] = [a 2b1b 2 ] and
the equation of the plane containing them is [r b1 b 2 ] = [a1 b1 b 2 ] or [r b1 b 2 ] = [a 2 b1 b 2 ] .
Note : Every pair of parallel lines is coplanar.
Two coplanar lines are either parallel or intersecting.
The three sides of a triangle are coplanar.
Important Tips
• Division by plane : The ratio in which the line segment PQ, joining P(x1, y1, z1) and Q(x2, y2, z2), is divided by plane
ax + by + cz + d =0 is = − ax1 + by1 + cz1 + d .
ax 2 + by 2 + cz 2 + d
• Division by co-ordinate planes : The ratio in which the line segment PQ, joining P(x1, y1, z1) and Q(x2, y2, z2) is divided by co-
ordinate planes are as follows :
(i) By yz-plane : –x1/x2 (ii) By zx-plane : –y1/y2 (ii) By xy-plane : –z1/z2
Example: 38 The xy-plane divides the line joining the points (–1, 3, 4) and (2, –5, 6) [Rajasthan PET 2000]
(a) Internally in the ratio 2 : 3 (b) Internally in the ratio 3 : 2
(c) Externally in the ratio 2 : 3 (d) Externally in the ratio 3 : 2
Solution: (c) Required ratio =− z1 = − 4 =− 2
Example: 39 z2 6 3
Solution: (c)
Example: 40 ∴ xy-plane divide externally in the ratio 2 : 3.
Solution: (d)
The ratio in which the plane x − 2y + 3z = 17 divides the line joining the point (–2, 4, 7) and (3, –5, 8) is [AISSE 1988]
Example: 41
Solution: (c) (a) 10 : 3 (b) 3 : 1 (c) 3 : 10 (d) 10 : 1
Example: 42 Required ratio = − ax1 + by1 + cz1 + d = − − 2 −8+ 21 − 17 = 6 = 3 .
Solution: (c) ax2 + by2 + cz2 + d 3 + 10 + 24 − 17 20 10
Example: 43
Solution: (b) The equation of the plane, which makes with co-ordinate axes a triangle with its centroid (α, β, γ), is [MP PET 2004]
Example: 44
(a) αx + βy + γz = 3 (b) x + y + z = 1 (c) αx + βy + γz = 1 (d) x + y + z = 3
αβγ αβγ
We know that xyz =1 ……(i)
a +b+c
Centroid a , b , c i.e. α = a / 3, β = b / 3,γ = c / 3 ⇒ a = 3α, b = 3β, c = 3γ
3 3 3
From equation (i), x + y + z =1
3α 3β 3γ
∴ x + y + z =3.
αβγ
The equation of plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1 is
[AISSE 1984; Tamilnadu (Engg.) 2002]
(a) 3x + 4y + 5z = 9 (b) 3x + 4y + 5z = 0 (c) 3x + 4y − 5z = 9 (d) None of these
We know that, equation of plane is a(x − x1) + b(y − y1) + c(z − z1) = 0
It passes through (2, 2, 1)
∴ a(x − 2) + b(y − 2) + c(z − 1) = 0 ……(i)
Plane (i) also passes through (9, 3, 6) and is perpendicular to the plane 2x + 6y + 6z = 1
∴ 7a + b + 5c = 0 ……(ii)
and 2a + 6b + 6c = 0 ……(iii)
abc or a bc
6 − 30 = 10 − 42 = 42 − 2 − 24 = − 32 = 40
or a b = c = k (say)
3 =4 −5
From equation (i), 3k(x − 2) + 4k(y − 2) + (−5)k(z − 1) = 0
Hence, 3x + 4y − 5z = 9 .
The equation of the plane containing the line r = a + kb and perpendicular to the plane r.n = q is
(a) (r − b).(n × a) = 0 (b) (r − a).(n × (a × b)) = 0 (c) (r − a).(n × b) = 0 (d) (r − b).(n × (a × b)) = 0
Since the required plane contains the line r = a + kb and is perpendicular to the plane r.n = q .
∴ It passes through the point a and parallel to vectors b and n. Hence, it is perpendicular to the vector N = n × b .
∴ Equation of the required plane is (r − a).N = 0 ⇒ (r − a).(n × b) = 0 .
The equation of the plane through the intersection of the planes x + 2y + 3z − 4 = 0 , 4 x + 3y + 2z + 1 = 0 and passing
through the origin will be [MP PET 1997; Kerala (Engg.) 2001; AISSE 1983]
(a) x + y + z = 0 (b) 17x + 14y + 11z = 0 (c) 7x + 4y + z = 0 (d) 17x + 14y + z = 0
Any plane through the given planes is (x + 2y + 3z − 4) + k(4 x + 3y + 2z + 1) = 0
It passes through (0, 0, 0)
∴ −4+k =0 =k=4
∴ Required plane is (x + 2y + 3z − 4) + 4(4 x + 3y + 2z + 1) = 0 ⇒ 17x + 14y + 11z = 0 .
The vector equation of the plane passing through the origin and the line of intersection of plane r.a = λ and r.b = µ is
(a) r.(λa − µb) = 0 (b) r.(λb − µa) = 0 (c) r.(λa + µb) = 0 (d) r.(λb + µa) = 0
Solution: (b) The equation of a plane through the line of intersection of plane r.a = λ and r.b = µ can be written as
Example: 45
Solution: (a) r.(a + kb) = λ + kµ ……(i)
Example: 46
Solution: (c) This passes through the origin, therefore putting the value of k in (i),
Example: 47 r(µa − λb) = 0 ⇒ r.(λb − µa) = 0 .
Solution: (a)
Angle between two planes x + 2y + 2z = 3 and −5x + 3y + 4z = 9 is [IIT Screening 2004]
Example: 48
Solution: (c) (a) cos−1 32 (b) cos−1 19 2 (c) cos−1 92 (d) cos−1 32
Example: 49 10 30 20 5
Solution: (b)
We know that, cos θ = a1a2 + b1b2 + c1c2 = 1(−5) + 2(3) + 2(4) = 9 = 32
1 + 4 + 4 25 + 9 + 16 3.5 10
a12 + b12 + c12 a22 + b22 + c22 2
i.e. θ = cos−1 32 .
10
Distance between two parallel planes 2x + y + 2z = 8 and 4 x + 2y + 4z + 5 = 0 is [AIEEE 2004]
(a) 9 (b) 5 (c) 7 (d) 3
2 2 2 2
We have 2x + y + 2z − 8 = 0 ……(i)
and 4 x + 2y + 4z + 5 = 0 or 2x + y + 2z + 5 / 2 = 0 ……(ii)
Distance between the planes = (5 / 2) + 8 = 21 = 7 .
4 +1+ 4 2.3 2
A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(–1, 1, 2). Then the angle between the faces OAB and
ABC will be
[MNR 1994; UPSEAT 2000; AIEEE 2003]
(a) cos −1 19 (b) cos−1 17 (c) 30° (d) 90°
35 31
Angle between two plane faces is equal to the angle between the normals n1 and n2 to the planes. n1 , the normal to the
i jk
face OAB is given by OA× OB = 1 2 1 = 5i − j − 3k ……(i)
21 3
n2 , the normal to the face ABC, is given by AB × AC .
i jk
n2 = 1 − 1 2 = i − 5j − 3k ……(ii)
−2 −1 1
If θ be the angle between n1 and n2 , Then cos θ = | n1.n2 | = 5.1 + 5 + 9
n1 || n 35 35
2
cos θ 19 ⇒ θ = cos−1 19 .
= 35 35
The distance of the point (2, 1, –1) from the plane x − 2y + 4z = 9 is [Kerala (Engg.) 2001]
(a) 13 (b) 13 (c) 13 (d) 13
21 21 21 21
Distance of the plane from (2, 1, –1) = 2 − 2(1) + 4(−1) − 9 = 13 .
1 + 4 + 16 21
A unit vector perpendicular to plane determined by the points P(1, –1, 2), Q(2, 0, –1) and R(0, 2, 1) is [IIT 1994]
(a) 2i − j + k (b) 2i + j + k (c) −2i + j + k (d) 2i + j − k
6 6 66
We know that, PQ × PR
| PQ × PR|
PQ = i + j − 3k , PR = −i + 3j − k
i jk
PQ × PR = 1 1 − 3 = 8i + 4j + 4k and| PQ × PR|= 4 6
−1 3 −1
Hence, the unit vector is 4(2i + j + k) i.e. 2i + j + k .
46 6
Example: 50 The perpendicular distance from origin to the plane through the point (2, 3, –1) and perpendicular to vector 3i − 4j + 7k
Solution: (a)
Example: 51 is
Solution: (c)
(a) 13 (b) − 13 (c) 13 (d) None of these
Example: 52 74 74
Solution: (b)
We know, the equation of the plane is (r − a).n = 0
Example: 53
Solution: (a) or (r − (2i + 3j − k)).(3i − 4j + 7k) = 0 ⇒ (xi + yj + zk − 2i − 3j + k).(3i − 4j + 7k) = 0 ⇒ 3x − 4y + 7z + 13 = 0
Hence, perpendicular distance of the plane from origin = 13 = 13 .
32 + (−4)2 + 72 74
If P = (0, 1, 0), Q =(0, 0, 1), then projection of PQ on the plane x + y + z = 3 is [EAMCET 2002]
(a) 3 (b) 3 (c) 2 (d) 2
Given plane is x + y + z − 3 = 0 . From point P and Q draw PM and QN perpendicular on the given plane and QR ⊥ MP.
| MP |= 0 + 1 + 0 − 3 = 2 3 P(0, 1, 0)
12 + 12 + 12 R
| NQ|= 2 (0,0,1) Q
3
| PQ|= (0 − 0)2 + (0 − 1)2 + (1 − 0)2 = 2
| RP |=| MP |−| MR|=| MP |−| NQ|= 0 (i.e. R and P are the same point) NM
∴ | NM |=|QR|= PQ2 − RP 2 = ( 2)2 − 0 = 2
The reflection of the point (2, –1, 3) in the plane 3x − 2y − z = 9 is [AMU 1995]
(a) 26 , 15 , 17 (b) 26 , −15 , 17 (c) 15 , 26 , −17 (d) 26 , 17 , −15
7 7 7 7 7 7 7 7 7 7 7 7
Let P be the point (2, –1, 3) and Q be its reflection in the given plane.
Then, PQ is perpendicular to the given plane
Hence, d.r.’s of PQ are 3, –2, 1 and consequently, equations of PQ are x − 2 = y +1 = z−3
3 −2 −1
Any point on this line is (3r + 2, − 2r − 1, − r + 3)
Let this point be Q. Then midpoint of PQ = 3r +2 + 2 , −2r −1 − 1 , −r +3 + 3 = 3r + 4 , − r − 1, −r + 6
2 2 2 2 2
This point lies in given plane i.e. 3 3r + 4 − 2(−r − 1) − −r + 6 = 9 ⇒ 9r + 12 + 4r + 4 + r − 6 = 9 ⇒ 14r =8 ⇒ r = 4
2 2 7
Hence, the required point Q is 3 4 + 2, − 2 4 − 1, −4 + 3 = 26 , − 15 , 17 .
7 7 7 7 7 7
A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the plane
determined by the vectors i – j, i + k. The angle between a and the vector i – 2j + 2k is [IIT 1996]
(a) π or 3π (b) 2π or 3π (c) π or 3π (d) None of these
4 4 4 4 2 2
Equation of plane containing i and i + j is
[r − i, i, i + j] = 0 ⇒ (r − i).[i × (i + j)] = 0 ⇒ [(x − 1) i + yj + zk].k = 0 ⇒ z = 0 ……(i)
Equation of plane containing i – j and i + k is …… (ii)
⇒ [r − (i − j) i − j i + k] = 0 ⇒ (r − i + j)[(i − j)× (i + k)] = 0 ⇒ x + y − z = 0
Let a = a1i + a2 j + a3 k . Since a is parallel to (i) and (ii)
a3 = 0 , a1 + a2 − a3 = 0 ⇒ a1 = −a2 , a3 = 0
Thus a vector in the direction of a is u = i − j . If θ is the angle between a and i − 2j + 2k .
Then cos θ = ± 1(1) + (−1)(−2) = ± 3 ⇒ cos θ = ± 1 ⇒ θ = π / 4 or 3π / 4
1+1 1+4+4 2. 3 2
Example: 54 The d.r.’s of normal to the plane through (1, 0, 0) and (0, 1, 0) which makes an angle π / 4 with plane x + y = 3 , are
Solution: (b)
[AIEEE 2002]
(a) 1, 2,1 (b) 1,1, 2 (c) 1, 1, 2 (d) 2,1,1
Let d.r.’s of normal to plane (a, b, c) ……(i)
a(x − 1) + b(y − 0) + c(z − 0) = 0
It is passes through (0, 1, 0). ∴ a + b = 0 ⇒ b = a . D.r.’s of normal is (a, a, c) and d.r.’s of given plane is (1, 1, 0)
∴ cos π / 4 = a+a+0 ⇒ 4a2 = 2a2 + c2 ⇒ 2a = c
a2 + a2 + c2 2
Then, d.r.’s of normal (a, a, 2a) or (1,1, 2) .
Line and plane
Equation of plane through a given line.
(1) If equation of the line is given in symmetrical form as x − x1 = y − y1 = z − z1 , then equation of plane is
l m n
a(x − x1) + b(y − y1) + c(z − z1) = 0 …… (i)
where a, b, c are given by al + bm + cn = 0 ……(ii)
(2) If equation of line is given in general form as a1 x + b1y + c1z + d1 = 0 = a2 x + b2y + c 2z + d2 , then the
equation of plane passing through this line is (a1 x + b1y + c1z + d1) + λ(a2 x + b2y + c 2z + d2 ) = 0 .
(3) Equation of plane through a given line parallel to another line : Let the d.c.’s of the other line be
l2,m2,n2 . Then, since the plane is parallel to the given line, normal is perpendicular.
∴ al2 + bm2 + cn2 = 0 ……(iii)
Hence, the plane from (i), (ii) and (iii) is x − x1 y − y1 z − z1
l1 m1 n1 = 0 .
l2 m2 n2
Transformation from unsymmetric form of the equation of line to the symmetric form.
If P ≡ a1 x + b1y + c1z + d1 = 0 and Q ≡ a2 x + b2y + c 2z + d2 = 0 are equations of two non-parallel planes,
then these two equations taken together represent a line. Thus the equation of straight line can be written as
P = 0 = Q . This form is called unsymmetrical form of a line.
To transform the equations to symmetrical form, we have to find the d.r.’s of line and co-ordinates of a point
on the line.
Intersection point of a line and plane.
To find the point of intersection of the line x − x1 = y − y1 = z − z1 and the plane ax + by + cz + d = 0 .
l m n
The co-ordinates of any point on the line
x − x1 = y − y1 = z − z1 are given by
l m n
P
x − x1 = y − y1 = z − z1 = r (say) or (x1 + lr, y1 + mr, z1 + nr) .....(i)
l m n (x1+lr, y1+mr, z1+nr)
ax+by+cz+d=0
If it lies on the plane ax + by + cz + d = 0 , then
a(x1 + lr) + b(y1 + mr) + c(z1 + n r) + d = 0 ⇒ (ax1 + by1 + cz1 + d) + r(al + bm + cn) = 0
∴ r = − (ax 1 + by1 + cz1 + d) .
al + bm + cn
Substituting the value of r in (i), we obtain the co-ordinates of the required point of intersection.
Algorithm for finding the point of intersection of a line and a plane
Step I : Write the co-ordinates of any point on the line in terms of some parameters r (say).
Step II : Substitute these co-ordinates in the equation of the plane to obtain the value of r.
Step III : Put the value of r in the co-ordinates of the point in step I.
Angle between line and plane.
(1) Cartesian form : The angle θ between the line x −α = y−β = z −γ , and the plane
l m n
ax + by + cz + d = 0 , is given by sinθ = al + bm + cn .
(a 2 + b 2 + c 2 ) (l 2 + m2 + n2 )
(i) The line is perpendicular to the plane if and only if a = b = c .
l m n
(ii) The line is parallel to the plane if and only if al + bm + cn = 0 .
(iii) The line lies in the plane if and only if al + bm + cn = 0 and aα + bβ + cγ + d = 0 .
(2) Vector form : If θ is the angle between a line r = (a + λb) and the plane r.n = d , then sinθ = | b.n | .
b|| n
(i) Condition of perpendicularity : If the line is perpendicular to the plane, then it is parallel to the normal
to the plane. Therefore b and n are parallel. n (π/2) –θ
So, b × n = 0 or b = λn for some scalar λ. r=a+λb
θ
r.n.=d
(ii) Condition of parallelism : If the line is parallel to the plane, then it is perpendicular to the normal to the
plane. Therefore b and n are perpendicular. So, b.n = 0.
(iii) If the line r = a + λb lies in the plane r.n = d, then (i) b.n = 0 and (ii) a.n = d.
Projection of a line on a plane.
If P be the point of intersection of given line and plane and Q be the foot of the perpendicular from any point
on the line to the plane then PQ is called the projection of given line on the given plane.
Image of line about a plane : Let line is x − x1 = y − y1 = z − z1 , plane is a2x + b2y + c2z + d = 0 .
a1 b1 c1
Find point of intersection (say P) of line and plane. Find image (say Q) of point (x1, y1, z1) about the plane.
Line PQ is the reflected line.
Example: 55 The sine of angle between the straight line x−2 = y−3 = z−4 and the plane 2x − 2y + z = 5 is
Solution: (b) 3 4 5
Example: 56
Solution: (a) [Kurukshetra CEE 1995, 2001; DCE 2000]
Example: 57 (a) 23 (b) 2 (c) 4 (d) 10
Solution: (c) 5 10 52 65
Example: 58 We know that sinθ = al + bm + cn
Solution: (a)
a2 + b2 + c2 l2 + m2 + n2
sin θ = 3(2) + 4(−2) + 5(1) = 3
9 + 16 + 25 4 + 4 + 1 5 2.3
Hence, sin θ = 2
10
Value of k such that the line x −1 = y −1 = z−k is perpendicular to normal to the plane r(2i + 3j + 4k) = 0 is
2 3 k
[Pb. CET 2001]
(a) − 13 (b) − 17 (c) 4 (d) None of these
4 4
We have, x − 1 = y −1 = z−k
2 3 k
or vector form of equation of line is r = (i + j + kk) + λ(2i + 3j + kk) i.e. b = 2i + 3j + kk and normal to the plane,
n = 2i + 3j + 4k .
Given that, b.n = 0
⇒ (2i + 3j + kk).(2i + 3j + 4k) = 0
⇒ 4 + 9 + 4k = 0 ⇒ k = −13 / 4 .
The equation of line of intersection of the planes 4 x + 4y − 5z = 12 , 8x + 12y − 13z = 32 can be written as [MP PET 2004]
(a) x = y −1 = z −2 (b) x = y = z −2 (c) x −1 = y−2 = z (d) x −1 = y−2 = z
2 3 4 2 3 4 2 3 4 2 −3 4
Let equation of line x − x1 = y − y1 = z − z1 ……(i)
l m n
We have 4 x + 4y − 5z = 12 ……(ii) and 8x + 12y − 13z = 32 ……(iii)
Let z = 0. Now putting z = 0 in (ii) and (iii),
we get, 4x + 4y = 12 , 8x + 12y = 32 , on solving these equations, we get x = 1, y = 2 .
Equation of line passing through (1, 2, 0) is x −1 = y−2 = z−0
l m n
From equation (i) and (ii),
4l + 4m − 5n = 0 and 8l + 12m − 13n = 0
⇒ l = m = n i.e. l = m = n . Hence, equation of line is x −1 = y − 2 = z .
8 12 16 2 3 4 2 3 4
The equation of the plane containing the two lines x −1 = y +1 = z and x = y−2 = z +1 is [MP PET 2000]
2 −1 3 2 −1 −3
(a) 8x + y − 5z − 7 = 0 (b) 8x + y + 5z − 7 = 0 (c) 8x − y − 5z − 7 = 0 (d) None of these
Any plane through the first line may be written as ……(i)
a(x − 1) + b(y + 1) + c(z) = 0
where, 2a − b + 3c = 0 ……(ii)
It will pass through the second line, if the point (0, 2, –1) on the second line also lies on (i)
i.e. if a(0 − 1) + b(2 + 1) + c(−1) = 0 , i.e., −a + 3b − c = 0 ……(iii)
Solving (ii) and (iii), we get a = b = c i.e. a = b = c
−8 −1 5 8 1 −5
∴ Required plane is 8(x − 1) + 1(y + 1) − 5(z) = 0 ⇒ 8x + y − 5z − 7 = 0 .
Example: 59 The plane which passes through the point (3, 2, 0) and the line x − 3 = y − 6 = z − 4 is [AIEEE 2002]
Solution: (a) 1 5 4
Example: 60 (a) x − y + z = 1 (b) x + y + z = 5 (c) x + 2y − z = 1 (d) 2x − y + z = 5
Solution: (d)
Example: 61 Any plane through the line x−3 = y−6 = z−4 is
Solution: (a) 1 5 4
Example: 62
Solution: (d) a(x − 3) + b(y − 6) + c(z − 4) = 0 ……(i)
where, a + 5b + 4c = 0 ……(ii)
Plane (i) passes through (3, 2, 0), if
a(3 − 3) + b(2 − 6) + c(0 − 4) = 0
−4b − 4c = 0 i.e. b + c = 0 ……(iii)
From equation (ii) and (iii), a + b = 0 . ∴ a = −b = c .
∴ Required plane is a(x − 3) − a(y − 6) + a(z − 4) = 0 i.e. x − y + z − 3 + 6 − 4 = 0 i.e. x − y + z = 1 .
x−3 y−6 z−4 x−3 y−6 z−4
Trick : 3 − 3 2 − 6 0 − 4 = 0 − 4 − 4 ⇒ x − y + z = 1 .
154 154
The distance of point (–1, –5, –10) from the point of intersection of the line x −2 = y +1 = z−2 and plane x−y+z =5
3 4 12
is [MP PET 2002]
(a) 10 (b) 8 (c) 21 (d) 13
Any point on the line x − 2 = y + 1 = z−2 =r is (3r + 2, 4r − 1,12r + 2)
3 4 12
This lies on x − y + z = 5 , then 3r + 2 − 4r + 1 + 12r + 2 = 5 i.e. r = 0.
∴ Point is (2, –1, 2). Its distance from (–1, –5, –10) is 9 + 16 + 144 = 13 .
The value of k such that x−4 = y−2 = z−k lies in the plane 2x − 4y + z = 7 is [IIT Screening 2003]
1 1 2
(a) 7 (b) –7 (c) No real value (d) 4
Given, point (4, 2, k) is on the line and it also passes through the plane 2x − 4y + z = 7 ⇒ 2(4) − 4(2) + k = 7 ⇒ k = 7 .
The distance between the line r = (i + j + 2k) + λ(2i + 5j + 3k) and the plane r.(2i + j − 3k) = 5 is
[Kurukshetra CEE 1996]
(a) 5 (b) 6 (c) 7 (d) 8
14 14 14 14
The given line is r = (i + j + 2k) + λ(2i + 5j + 3k)
a = i + j + 2k , b = 2i + 5j + 3k
Given plane, r.(2i + j − 3k) = 5 ⇒ r.n = p
Since b.n = 4 + 5 − 9 = 0
∴ The line is parallel to plane. Thus the distance between line and plane is equal to length of perpendicular from a point
a = i + j + 2k on line to given plane.
Hence, required distance = (i + j + 2k).(2i + j − 3k) − 5 = 2 + 1 − 6 − 5 = 8 .
4 +1+ 9 14 14
Sphere
A sphere is the locus of a point which moves in space in such a way that its distance from a fixed point always
remains constant.
The fixed point is called the centre and the constant distance is called the radius of the sphere.
P(r)
C(a)
General equation of sphere.
The general equation of a sphere is x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0 with centre (–u, –v, –w)
i.e. (–(1/2) coeff. of x, –(1/2) coeff. of y, –(1/2) coeff. of z) and, radius = u2 + v2 + w2 − d
From the above equation, we note the following characteristics of the equation of a sphere :
(i) It is a second degree equation in x, y, z;
(ii) The coefficients of x 2, y 2, z 2 are all equal;
(iii) The terms containing the products xy, yz and zx are absent.
Note : The equation x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0 represents,
(i) A real sphere, if u2 + v2 + w2 − d > 0 .
(ii) A point sphere, if u2 + v2 + w2 − d = 0 .
(iii) An imaginary sphere, if u2 + v2 + w2 − d < 0 .
Important Tips
• If u2 + v2 + w2 − d < 0 , then the radius of sphere is imaginary, whereas the centre is real. Such a sphere is called “pseudo-sphere” or a
“virtual sphere.
• The equation of the sphere contains four unknown constants u, v, w and d and therefore a sphere can be found to satisfy four conditions.
Equation in sphere in various forms.
(1) Equation of sphere with given centre and radius
(i) Cartesian form : The equation of a sphere with centre (a, b, c) and radius R is
(x − a)2 + (y − b)2 + (z − c)2 = R 2 ……(i)
If the centre is at the origin, then equation (i) takes the form x 2 + y 2 + z 2 = R 2 ,
which is known as the standard form of the equation of the sphere.
(ii) Vector form : The equation of sphere with centre at C(c) and radius ‘a’ is | r − c|= a .
(2) Diameter form of the equation of a sphere
(i) Cartesian form : If (x1, y1, z1) and (x 2, y2, z2 ) are the co-ordinates of the extremities of a diameter of a
sphere, then its equation is (x − x1)(x − x 2 ) + (y − y1)(y − y2 ) + (z − z1)(z − z2 ) = 0 .
(ii) Vector form : If the position vectors of the extremities of a diameter of a sphere are a and b, then its
equation is (r − a).(r − b) = 0 or | r |2 −r.(a − b) + a.b = 0 .
Section of a sphere by a plane. C
PM
Consider a sphere intersected by a plane. The set of points common to
both sphere and plane is called a plane section of a sphere. The plane section of
a sphere is always a circle. The equations of the sphere and the plane taken
together represent the plane section.
Q
Let C be the centre of the sphere and M be the foot of the perpendicular from C on the plane. Then M is the
centre of the circle and radius of the circle is given by PM = CP 2 − CM 2
The centre M of the circle is the point of intersection of the plane and line CM which passes through C and is
perpendicular to the given plane.
Centre : The foot of the perpendicular from the centre of the sphere to the plane is the centre of the circle.
(radius of circle)2 = (radius of sphere)2 – (perpendicular from centre of spheres on the plane)2
Great circle : The section of a sphere by a plane through the centre of the sphere is a great circle. Its centre
and radius are the same as those of the given sphere.
Condition of tangency of a plane to a sphere.
A plane touches a given sphere if the perpendicular distance from the centre of the sphere to the plane is
equal to the radius of the sphere.
(1) Cartesian form : The plane lx + my + nz = p touches the sphere x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0 ,
if (ul + vm + wn − p)2 = (l 2 + m2 + n2 )(u2 + v2 + w2 − d)
(2) Vector form : The plane r.n =d touches the sphere | r − a|= R if | a.n − d | = R.
|n|
Important Tips
• Two spheres S1 and S2 with centres C1 and C2 and radii r1 and r2 respectively
(i) Do not meet and lies farther apart iff |C1C2 |> r1 + r2
(ii) Touch internally iff |C1C2 |=|r1 − r2 |
(iii) Touch externally iff |C1C2 |= r1 + r2
(iv) Cut in a circle iff |r1 − r2 |<|C1C2 |< r1 + r2
(v) One lies within the other if |C1C2 |<|r1 − r2 |.
When two spheres touch each other the common tangent plane is S1 − S2 = 0 and when they cut in a circle, the plane of the circle is
S1 − S2 = 0 ; coefficients of x2, y2, z2 being unity in both the cases.
• Let p be the length of perpendicular drawn from the centre of the sphere x2 + y2 + z2 = r2 to the plane Ax + By + Cz + D = 0 , then
(i) The plane cuts the sphere in a circle iff p < r and in this case, the radius of circle is r2 − p2 .
(ii) The plane touches the sphere iff p = r .
(iii) The plane does not meet the sphere iff p > r.
• Equation of concentric sphere : Any sphere concentric with the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 is
x2 + y2 + z2 + 2ux + 2vy + 2wz + λ = 0 , where λ is some real which makes it a sphere.
Intersection of straight line and a sphere.
Let the equations of the sphere and the straight line be x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 .....(i)
And x −α = y−β = z −γ =r (say) .....(ii)
l m n
Any point on the line (ii) is (α + lr, β + mr,γ + nr) .
If this point lies on the sphere (i) then we have,
(α + lr)2 + (β + mr)2 + (γ + nr)2 + 2u(α + lr) + 2v(β + mr) + 2w(γ + nr) + d = 0
or, r 2[l 2 + m2 + n2] + 2r[l(u + α) + m(v + β )] + n(w + γ )] + (α 2 + β 2 + γ 2 + 2uα + 2vβ + 2wγ + d) = 0 ....(iii)
This is a quadratic equation in r and so gives two values of r and therefore the line (ii) meets the sphere (i)
in two points which may be real, coincident and imaginary, according as root of (iii) are so.
Note : If l, m, n are the actual d.c.’s of the line, then l 2 + m2 + n2 = 1 and then the equation (iii) can
be simplified.
Angle of intersection of two spheres.
The angle of intersection of two spheres is the angle between the tangent planes to them at their point of
intersection. As the radii of the spheres at this common point are normal to the tangent planes so this angle is also
equal to the angle between the radii of the spheres at their point of intersection.
If the angle of intersection of two spheres is a right angle, the spheres are said to be orthogonal.
Condition for orthogonality of two spheres
Let the equation of the two spheres be
x 2 + y2 + z 2 + 2ux + 2vy + 2wz + d = 0 .....(i)
and x 2 + y2 + z 2 + 2u′x + 2v′y + 2w′z + d′ = 0 .....(ii)
If the sphere (i) and (ii) cut orthogonally, then 2uu′ + 2vv′ + 2ww′ = d + d′, which is the required condition.
Note : If the spheres x 2 + y2 + z 2 = a2 and x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0 cut
orthogonally, then d = a2 . r1r2 .
Two spheres of radii r1 and r2 cut orthogonally, then the radius of the common circle is r12 + r22
Example: 63 The centre of sphere passing through four points (0, 0, 0), (0, 2, 0), (1, 0, 0) and (0, 0, 4) is [MP PET 2002]
Solution: (a)
(a) 1 , 1, 2 (b) − 1 , 1, 2 (c) 1 , 1, − 2 (d) 1, 1 , 2
Example: 64 2 2 2 2
Solution: (b)
Example: 65 Let the equation of sphere be x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0 [DCE 1998]
Solution: (a)
It passes through (0, 0, 0), ∴ d = 0
Also, It passes through (0, 2, 0) i.e., v = −1
Also, It passes through (1, 0, 0) i.e., u = −1 / 2
Also, it passes through (0, 0, 4) i.e., w = −2
∴ Centre (–u, –v, –w) = (1/2, 1, 1/2)
The equation | r |2 −r.(2i + 4j − 2k) − 10 = 0 represents a
(a) Plane (b) Sphere of radius 4 (c) Sphere of radius 3 (d) None of these
The given equation is | r |2 −r(2i + 4j − 2k) − 10 = 0
⇒ x 2 + y 2 + z 2 − 2x − 4y + 2z − 10 = 0 ,
which is the equation of sphere, whose centre is (1, 2 –1) and radius = 1 + 4 + 1 + 10 = 4 .
The intersection of the spheres x 2 + y 2 + z 2 + 7x − 2y − z = 13 and x 2 + y 2 + z 2 − 3x + 3y + 4z = 8 is the same as the
intersection of one of the sphere and the plane [AIEEE 2004]
(a) 2x − y − z = 1 (b) x − 2y − z = 1 (c) x − y − 2z = 1 (d) x − y − z = 1
We have the spheres x 2 + y 2 + z 2 + 7x − 2y − z − 13 = 0 and x 2 + y 2 + z 2 − 3x + 3y + 4z − 8 = 0
Required plane is S1 − S2 = 0
∴ (7x + 3x) − (2y + 3y) − (z + 4z) − 5 = 0
i.e. 10x − 5y + (−5z) − 5 = 0 ⇒ 2x − y − z = 1 .
Example: 66 The radius of the circle in which the sphere x2 + y2 + z2 + 2x − 2y − 4z − 19 = 0 is cut by the plane x + 2y + 2z + 7 = 0 is
Solution: (c)
[AIEEE 2003]
(a) 1 (b) 2 (c) 3 (d) 4
For sphere x 2 + y 2 + z 2 + 2x − 2y − 4z − 19 = 0 , Centre O is (–1, 1, 2) and radius = 1 + 1 + 4 + 19 = 5 ,
Now, OL = length of perpendicular from O to plane x + 2y + 2z + 7 = 0 is
= −1 + 2 + 4 +7 = 12 = 4 , i.e. OL = 4 .
1+ 4 + 4 3
O (–1, 1, 2)
In ∆OLB, LB = OB2 − OL2 = 25 − 16 = 3 . 5
A LB
Example: 67 The radius of circular section of the sphere | r |= 5 by the plane r. (i + j + k) = 4 3 is [DCE 1999; AMU 1991]
Solution: (b)
(a) 2 (b) 3 (c) 4 (d) 6
Example: 68
Solution: (c) Radius of the sphere =5
Given plane is x + y + z − 4 3 = 0
Length of the perpendicular from the centre (0, 0, 0) of the sphere to the plane = 43 = 4
1+1+1
Hence, radius of circular section = 25 − 16 = 3 .
The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2 + y2 + z2 + 4 x − 2y − 6z = 155 is [AIEEE 2003]
(a) 26 (b) 11 4 (c) 13 (d) 39
13
Centre of sphere is (–2, 1, 3)
Radius of sphere is 4 + 1 + 9 + 155 = 13
Distance of centre from plane = −24 + 4 + 9 − 327 = 338
144 + 16 + 9 13
∴ Plane cuts the sphere and hence S.D. . = 338 − 13 = 169 = 13 .
13 13
Statistics
Introduction.
Statics is that branch of mechanics which deals with the study of the system of forces in equilibrium.
Matter : Matter is anything which can be perceived by our senses of which can exert, or be acted on, by forces.
Force : Force is anything which changes, or tends to change, the state of rest, or uniform motion, of a body.
To specify a force completely four things are necessary they are magnitude, direction, sense and point of
application. Force is a vector quantity.
Parallelogram law of Forces.
If two forces, acting at a point, be represented in magnitude and direction by the two sides of a parallelogram
drawn from one of its angular points, their resultant is represented both in magnitude and direction of the
parallelogram drawn through that point.
If OA and OB represent the forces P and Q acting at a point O and inclined to each other at an angle α. If R is
the resultant of these forces represented by the diagonal OC of the parallelogram OACB and R makes an angle θ
with P i.e. ∠COA = θ , then R2 = P2 + Q2 + 2PQ cosα and tanθ = Q sinα
P + Q cosα
The angle θ1 which the resultant R makes with the direction of the force Q is given by
θ1 = tan −1 Q P sinα
+ P cosα
B C
Case (i) : If P = Q R
Q
∴ R = 2P cos(α / 2) and tanθ = tan(α / 2) or θ = α / 2 α
θ1 P AD
Case (ii) : If α = 90° , i.e. forces are perpendicular α
∴R = P2 + Q2 and tanθ = Q θ
P
O
Case (iii) : If α = 0° , i.e. forces act in the same direction
∴ Rmax = P + Q
Case (iv) : If α = 180° , i.e. forces act in opposite direction
∴ Rmin = P − Q
Note : The resultant of two forces is closer to the larger force.
The resultant of two equal forces of magnitude P acting at an angle α is 2P cos α and it bisects the
2
angle between the forces.
If the resultant R of two forces P and Q acting at an angle α makes an angle θ with the direction of
P, then sinθ = Q sinα and cosθ = P + Q cosα
R R
B C
If the resultant R of the forces P and Q acting at an angle α makes R
F2
an angle θ with the direction of the force Q, then β F1 A
α
sinθ = P sinα and cosθ = Q + P sinα
R R O
Component of a force in two directions : The component of a force R in two directions making
angles α and β with the line of action of R on and opposite sides of it are
F1 = OC. sin β = R sin β ) and F2 = OC. sin α = R. sin α
sin(α + β ) sin(α + β sin(α + β ) sin(α + β )
λ-µ theorem : The resultant of two forces acting at a point O in directions OA and OB represented in
magnitudes by λ.OA and µ.OB respectively is represented by (λ + µ)OC , where C is a point in AB such that
λ.CA = µ.CB A
λ OA C
OB
µ OB
Important Tips
• The forces P, Q, R act along the sides BC, CA, AB of ∆ABC.
Their resultant passes through.
(a) Incentre, if P + Q + R = 0 (b) Circumcentre, if P cos A + Q cos B + R cos C = 0
(c) Orthocentre, if P sec A + Q sec B + R sec C = 0 (d) Centroid, if P cosec A + Q cosec B + R cosec C = 0
or P = Q = R
a b c
Example: 1 Forces M and N acting at a point O make an angle 150°. Their resultant acts at O has magnitude 2 units and is
Solution: (a) perpendicular to M. Then, in the same unit, the magnitudes of M and N are [BIT Ranchi 1993]
Example: 2 (a) 2 3,4 (b) 3 ,2
Solution: (a) (c) 3, 4 2
Example: 3
(d) 4. 5
We have, 22 = M 2 + N 2 + 2MN cos150° ⇒ 4 = M 2 + N 2 − 3MN .....(i)
and, tan π = M sin150° ⇒ M + N cos150° = 0
2 M + N cos150°
⇒ M−N 3 =0⇒ M = N3 .....(ii)
2 2
Solving (i) and (ii), we get M = 2 3 and N = 4 .
If the resultant of two forces of magnitude P and 2P is perpendicular to P, then the angle between the forces is
[Roorkee 1997]
(a) 2π/3 (b) 3π/4 (c) 4π/5 (d) 5π/6
Let the angle between the forces P and 2P be α. Since the resultant of P and 2P is perpendicular to P. Therefore,
tanπ /2= 2P sinα ⇒ P + 2P cosα = 0 ⇒ cos α = −1 ⇒ α = 2π
P + 2P cosα 2 3
If the line of action of the resultant of two forces P and Q divides the angle between them in the ratio 1 : 2, then the
magnitude of the resultant is [Roorkee 1993]
(a) P2 + Q2 (b) P2 + Q2 (c) P2 − Q2 (d) P2 − Q2
P Q P Q
Solution: (d) Let 3θ be the angle between the forces P and Q. It is given that the resultant R of P and Q divides the angle between them in
Example: 4 the ratio 1 : 2. This means that the resultant makes an angle θ with the direction of P and angle 2θ with the direction of Q.
Solution: (b)
Therefore, P = R sin 2θ and θ = R sinθ
sin 3θ sin 3θ
⇒ P = sin 2θ = 2 cosθ .....(i)
Q sinθ
Also Q= R sinθ ⇒Q= R
sin 3θ 3 − 4 sin 2 θ
⇒ R = 3 − 4 sin 2 θ ⇒ R = −1 + 4 cos 2 θ ⇒ R +1 = (2 cosθ )2 .....(ii)
Q Q Q
P 2 R R P2 − Q2 P2 − Q2
Q Q Q Q2 Q
From (i) and (ii), we get, = +1⇒ = ⇒ R =
Two forces X and Y have a resultant F and the resolved part of F in the direction of X is of magnitude Y. Then the angle
between the forces is
(a) sin−1 X (b) 2 sin−1 X (c) 4 sin−1 X (d) None of these
2Y 2Y 2Y
Let OA and OB represent two forces X and Y respectively. Let α be the angle between them and θ, the angle which the
resultant F (represented by OC) makes with OA.
Now, resolved part of F along OA.
F cosθ = OC × OD = OD = OA + AD = OA + AC cosα = X + Y cos α B C
OC
But resolved part of F along OA is given to by Y.
∴ Y = X + Y cos α or Y(1 − cos α) = X ⇒ Y.2 sin2 α = X, ∴ sin2 α / 2 = X Y F Y
2 2Y X
α α
i.e., sin α = X or α = sin−1 X θ A D
2 2Y 2 2Y
O
Thus, α = 2 sin−1 X
2Y
Example: 5 The greatest and least magnitude of the resultant of two forces of constant magnitude are F and G. When the forces act an
angle 2α, the resultant in magnitudes is equal to [UPSEAT 2001]
(a) F 2 cos 2 α + G2 sin 2 α (b) F 2 sinα + G2 cos 2 α (c) F 2 + G2 (d) F 2 − G2
Solution: (a) Greatest resultant = F = A + B
Least resultant = G = A − B
On solving, we get A= F +G , B= (F − G)
2 2
where A and B act an angle 2α, the resultant
R = A2 + B2 + 2AB cos 2α ⇒ R = F 2 cos 2 α + G2 sin 2 α
Triangle law of Forces .
If three forces, acting at a point, be represented in magnitude and direction by the sides of a triangle, taken in
order, they will be in equilibrium. AR
Here AB = P, BC = Q, CA = R
In triangle ABC, we have AB + BC + CA = 0 PR Q
⇒ P+Q+R=0 B Q CP
Hence the forces P,Q,R are in equilibrium.
Converse : If three forces acting at a point are in equilibrium, then they can be represented in magnitude and
direction by the sides of a triangle, taken in order.
Polygon law of Forces .
If any number of forces acting on a particle be represented in magnitude and direction by the sides of a
polygon taken in order, the forces shall be in equilibrium.
P4 A3
A4 P3
A2
P5 P2
A P1 A1
Example: 6 D and E are the mid-points of the sides AB and AC respectively of a ∆ABC. The resultant of the forces is represented by
Solution: (d)
BE and DC is
(a) 3 AC (b) 3 CA (c) 3 AB (d) 3 BC
2 2 2 2
We have, A
( ) ( )BE + DC = BC + CE + DB + BC
= 2BC + 1 CA + 1 AB = 2BC + 1 CA + AB DE
2 2 2
= 2BC + 1 CB = 2BC − 1 BC = 3 BC
2 2 2
BC
Example: 7 ABCDE is pentagon. Forces acting on a particle are represented in magnitude and direction by AB, BC, CD,2DE, AD, and
AE . Their resultant is given by [Roorkee 1994]
(a) AE (b) 2AB (c) 3AE (d) 4 AE
Solution: (c) ( ) ( ) ( )We have, AB + BC + CD + 2DE + AD + AE = AB + BC + CD + DE + AD + DE + AE
( )= AC + CE + AE + AE = AE + AE + AE = 3AE .
Lami's Theorem .
If three forces acting at a point be in equilibrium, each force is proportional to the sine of the angle between
the other two. Thus if the forces are P, Q and R; α,β,γ be the angles between Q and R, R and P, P and Q
respectively. If the forces are in equilibrium, we have, P = Q = R . PR
sinα sin β sin γ β
The converse of this theorem is also true.
α
γ
Q
Example: 8 A horizontal force F is applied to a small object P of mass m on a smooth plane inclined to the horizon at an angle θ. If F is
just enough to keep P in equilibrium, then F = [BIT Ranchi 1993]
(a) mg cos2 θ (b) mg sin2 θ (c) mg cosθ (d) mg tanθ
Solution: (d) By applying Lami's theorem at P, we have RB
RF mg
sin 90° = sin(180° −θ ) = sin(90° + θ )
P
⇒ R = F = mg ⇒F = mg tanθ 90–θmg F
1 sin θ cos θ θ C
A
Example: 9 A kite of weight W is flying with its string along a straight line. If the ratios of the resultant air pressure R to the tension T in
Solution: (b)
the string and to the weight of the kite are 2 and ( 3 + 1) respectively, then [Roorkee 1990]
Example: 10
Solution: (d) (a) T = ( 6 + 2)W (b) R = ( 3 + 1)W (c) T = 1 ( 6− 2)W (d) R = ( 3 − 1)W
2
From Lami's theorem, RQ
RT W
sin(θ + φ) = sin(180o −θ ) = sin(180o − φ)
⇒ R = T = W .....(i) φT
sin(θ + φ) sin θ sin φ .....(iii) θ
P
Given, R = 2 .....(ii) and R = 3 +1 W
T W
R 3 +1
2
Dividing (iii) by (ii), we get W =
R
T
⇒ T = 3 +1 ⇒ T = 3 +1W = 1 ( 6+ 2)W ⇒ R = T 2 = 2 ( 3 + 1)W = ( 3 + 1)W
W 2 2 2 2
Three forces P, Q R are acting at a point in a plane. The angles between P and Q and Q and R are 150° and 120°
respectively, then for equilibrium, forces P, Q, R are in the ratio [MNR 1991; UPSEAT 2000]
(a) 1 : 2 : 3 (b) 1 : 2 : 3 (c) 3:2:1 (d) 3 : 2 : 1
Clearly, the angle between P and R is 360° − (150° + 120°) = 90° . By Lami's theorem,
P = Q = R ⇒ P 2 = Q = R ⇒ P = Q = R
sin 120° sin 90° sin 150° 3/ 1 1/ 2 3 2 1
Parallel Forces.
(1) Like parallel forces : Two parallel forces are said to be like parallel forces when they act in the same direction.
The resultant R of two like parallel forces P and Q is equal in magnitude of the sum of the magnitude of forces
and R acts in the same direction as the forces P and Q and at the point on the line segment joining the point of
action P and Q, which divides it in the ratio Q : P internally.
P RQ
A CB
(2) Two unlike parallel forces : Two parallel forces are said to be unlike if they act in opposite directions.
If P and Q be two unlike parallel force acting at A and B and P is greater in P
magnitude than Q. Then their resultant R acts in the same direction as P and acts R
at a point C on BA produced. Such that R = P − Q and P.CA = Q.CB B
Then in this case C divides BA externally in the inverse ratio of the forces, A
C
P = Q = P −Q = R
CB CA CB − CA AB
Q
Important Tips
• If three like parallel forces P, Q, R act at the vertices A, B, C repectively of a triangle ABC, then their resultant act at the
(i) Incentre of ∆ABC, if P Q R
a =b= c
(ii) Circumcentre of ∆ABC, if P = Q = R
sin 2A sin 2B sin 2C
(iii) Orthocentre of ∆ABC, if P = Q = R
tan A tan B tan C
(iv) Centroid of ∆ABC, if P = Q = R.
Example: 11 Three like parallel forces P, Q, R act at the corner points of a triangle ABC. Their resultant passes through the
circumcentre,if [Rookee 1995]
(a) P = Q = R (b) P = Q = R (c) P + Q + R = 0 (d) None of these
a b c
Solution: (c) Since the resultant passes through the circumcentre of ∆ABC, therefore, the algebraic sum of the moments about it, is zero.
Example: 12
Hence, P + Q + R = 0.
P and Q are like parallel forces. If P is moved parallel to itself through a distance x, then the resultant of P and Q moves
through a distance. [Rookee 1995]
(a) Px (b) Px (c) Px (d) None of these
P+Q P −Q P + 2Q
Solution: (a) Let the parallel forces P and Q act at A and B respectively. Suppose the resultant P + Q acts at C.
Then, AC = AB Q .....(i)
P+Q
A A′ C C′ B
If P is moved parallel to itself through a distance x i.e. at A'. x
Suppose the resultant now acts at C'. Then ,
A' C' = A' B Q ⇒ A' C' = AB − x Q .....(ii) PP P+Q P+Q Q
P+Q P+Q
Now CC' = AC'− AC = AA'+ A' C'− AC
⇒ CC' = x + AB − x Q − AB Q ⇒ CC' = x − Qx ⇒ CC' = Px
P+Q P+Q P+Q P+Q
Moment. O
The moment of a force about a point O is given in magnitude by the
product of the forces and the perpendicular distance of O from the line of action
of the force.
If F be a force acting a point A of a rigid body along the line AB and OM (= p) P
be the perpendicular distance of the fixed point O from AB, then the moment of
force about O = F.p = AB × OM = 2 1 (AB × OM) = 2(area of ∆AOB) F B
2 AM
The S.I. unit of moment is Newton-meter (N-m).
(1) Sign of the moment : The moment of a force about a point measures the tendency of the force to cause
rotation about that point. The tendency of the force F1 is to turn the lamina in the clockwise
direction and of the force F2 is in the anticlockwise direction. F2
The usual convention is to regard the moment which is anticlockwise direction as
positive and that in the clockwise direction as negative.
(2) Varignon's theorem : The algebraic sum of the moments of any two coplanar forces O
about any point in their plane is equal to the moment of their resultant about the same point.
Note : Thy algebraic sum of the moments of any two forces about any point on the F1
line of action of their resultant is zero.
Conversely, if the algebraic sum of the moments of any two coplanar forces, which are not in
equilibrium, about any point in their plane is zero, their resultant passes through the point.
If a body, having one point fixed, is acted upon by two forces and is at rest. Then the moments of
the two forces about the fixed point are equal and opposite.
Couples.
Two equal unlike parallel forces which do not have the same line of action, are said to form a couple.
Example : Couples have to be applied in order to wind a watch, to drive a P
gimlet, to push a cork screw in a cork or to draw circles by means of pair of
compasses. A pB
(1) Arm of the couple : The perpendicular distance between the lines of
action of the forces forming the couple is known as the arm of the couple.
(2) Moment of couple : The moment of a couple is obtained in P
magnitude by multiplying the magnitude of one of the forces forming the couple
and perpendicular distance between the lines of action of the force. The perpendicular distance between the forces
is called the arm of the couple. The moment of the couple is regarded as positive or negative according as it has a
tendency to turn the body in the anticlockwise or clockwise direction.
Moment of a couple = Force × Arm of the couple = P.p
(3) Sign of the moment of a couple : The moment of a couple is taken with positive or negative sign
according as it has a tendency to turn the body in the anticlockwise or clockwise direction.
PP
AB AB
Positive couple Negative couple P
P
Note : A couple can not be balanced by a single force, but can be balanced by a couple of opposite
sign.
Triangle theorem of Couples.
If three forces acting on a body be represented in magnitude, direction and line of action by the sides of
triangle taken in order, then they are equivalent to a couple whose moment is represented by twice the area of
triangle.
Consider the force P along AE, Q along CA and R along AB. These forces are three concurrent forces acting
at A and represented in magnitude and direction by the sides BC, CA and AB of A P
∆ABC. So, by the triangle law of forces, they are in equilibrium. DP E
The remaining two forces P along AD and P along BC form a couple, RQ
whose moment is m = P.AL = BC.AL
Since 1 (BC.AL) = 2 1 area of the ∆ABC B L PC
2 2
∴ Moment = BC.AL = 2 (Area of ∆ABC)
Example: 13 A light rod AB of length 30 cm. rests on two pegs 15 cm. apart. At what distance from the end A the pegs should be placed
Solution: (c) so that the reaction of pegs may be equal when weight 5W and 3W are suspended from A and B respectively
Example: 14 (a) 1.75 cm., 15.75 cm. (b) 2.75 cm., 17.75 cm. (c) 3.75 cm., 18.75 cm. [Roorkee 1995, UPSEAT 2001]
Solution: (b)
(d) None of these
Example: 15
Let R, R be the reactions at the pegs P and Q such that AP = x
Resolving all forces vertically, we get R 15 R
R + R = 8W ⇒ R = 4W Ax B
Take moment of forces about A, we get
R.AP + R.AQ = 3W.AB
⇒ 4W.x + 4W.(x + 15) = 3W.30 5W 3W
⇒ x = 3.75cm
∴ AP = x = 3.75cm and AQ = 18.75cm
At what height from the base of a vertical pillar, a string of length 6 metres be tied, so that a man sitting on the ground and
pulling the other end of the string has to apply minimum force to overturn the pillar [Roorkee 1997, SCRA 2000]
(a) 1.5 metres (b) 3 2 metres (c) 3 3 metres (d) 4 2 metres
Let the string be tied at the point C of the vertical pillar, so that AC = x B
Now moment of F about A = F. AL
= F. AP sinθ 6L C
= F.6 cosθ sin θ F A
= 3 F sin 2θ
To overturn the pillar with maximum (fixed) force F, moment is maximum if θ
sin 2θ = 1 (max.) P
⇒ 2θ = 90°, i.e. θ = 45°
∴ AC = PC sin 45° = 6. 1 = 3 2
2
Two unlike parallel forces acting at points A and B form a couple of moment G. If their lines of action are turned through a
right angle, they form a couple of moment H. Show that when both act at right angles to AB, they form a couple of
moment.
(a) GH (b) G2 + H2 (c) G2 + H 2 (d) None of these
Solution: (c) We have, Pa = G and Pb = H .....(i) PB
Clearly, a 2 + b 2 = x 2
Example: 16
Solution: (b) ⇒x= G2 H2 [from (i)] PP x
P2 + P2 a PP
⇒ Px = G2 + H 2 b AP
Hence, required moment = G2 + H 2
The resultant of three forces represented in magnitude and direction by the sides of a triangle ABC taken in order with BC
= 5 cm, CA = 5 cm, and AB = 8 cm, is a couple of moment
(a) 12 units (b) 24 units (c) 36 units (d) 16 units
Resultant of three forces represented in magnitude and direction by the sides of a triangle taken in order is a couple of
moment equal to twice the area of triangle.
∴ the resultant is a couple of moment = 2 × (area of ∆ABC)
Here, a = 5 cm, b = 5 cm and c = 8 cm
∴ 2S = 5 + 5 + 8 ⇒ S = 9.
Area = S(S − a)(S − b)(S − c) = 9(9 − 5)(9 − 5)(9 − 8) = 12
∴ Required moment = 2 (12) = 24 units.
Equilibrium of Coplanar Forces.
(1) If three forces keep a body in equilibrium, they must be coplanar.
(2) If three forces acting in one plane upon a rigid body keep it in equilibrium, they must either meet in a
point or be parallel.
(3) When more than three forces acting on a rigid body, keep it in equilibrium, then it is not necessary that they
meet at a point. The system of forces will be in equilibrium if there is neither translatory motion nor rotatory motion.
i.e. X = 0, Y = 0, G = 0 or R = 0, G = 0.
(4) A system of coplanar forces acting upon a rigid body will be in equilibrium if the algebraic sum of their
resolved parts in any two mutually perpendicular directions vanish separately, and if the algebraic sum of their
moments about any point in their plane is zero. A
αβ
(5) A system of coplanar forces acting upon a rigid body will be in
equilibrium if the algebraic sum of the moments of the forces about each of
three non-collinear points is zero. B θ
m P nC
(6) Trigonometrical theorem : If P is any point on the base BC of ∆ABC
such that BP : CP = m : n.
Then, (i) (m + n)cotθ = m cotα − n cot β where ∠BAP = α, ∠CAP = β
(ii) (n + n)cotθ = n cot B − m cot C
Example: 17 Two smooth beads A and B, free to move on a vertical smooth circular wire, are connected by a string. Weights W1, W2
Solution: (a) and W are suspended from A, B and a point C of the string respectively.
In equilibrium, A and B are in a horizontal line. If ∠BAC = α and ∠ABC = β , then the ratio tanα : tan β is
(a) tanα = W − W1 + W2 (b) tanα = W + W1 − W2 (c) tanα = W + W1 + W2 [Roorkee 1996, UPSEAT 2001]
tan β W + W1 − W2 tan β W − W1 + W2 tan β W + W1 − W2
(d) None of these
Resolving forces horizontally and vertically at the points A, B and C respectively, we get
T cosα = R1 sinγ .....(i)
T1 sinα + W1 = R1 cos γ .....(ii)
T1 cos β = R2 sin γ .....(iii) R1 R2
T2 sin β + W2 = R2 cos γ .....(iv)
T1 cosα = T2 cos β .....(v) A αβ B
and T1 sinα + T2 sin β = W .....(vi) γ T2 γ
Using (v), from (i)and (ii), we get, R1 = R2 C
T1 O
W1 W W2
∴ From (ii) and (vi), we have
T1 sin α + W1 = T2 sin β + W2
or T1 sinα − T2 sin β = W2 − W1 .....(vii)
Adding and subtracting (vi) and (vii), we get
2T1 sinα = W + W2 − W1 ......(viii)
2T2 sin β = W − W2 + W1 ......(ix)
Dividing (viii) by (ix), we get
T1 . sin α = W − W1 + W2 or cos β . sin α = W − W1 + W2 (from (v)) or tanα = W − W1 + W2
T2 sin β W + W1 − W2 cos α sin β W + W1 − W2 tan β W + W1 − W2
Example: 18 A uniform beam of length 2a rests in equilibrium against a smooth vertical plane and over a smooth peg at a distance h
Solution: (a)
from the plane. If θ be the inclination of the beam to the vertical, then sin3θ is [MNR 1996]
Example: 19
Solution: (b) (a) h (b) h2 (c) a (d) a2
a a2 h h2
Let AB be a rod of length 2a and weight W. It rests against a smooth vertical wall at A and over peg C, at a distance h from
the wall. The rod is in equilibrium under the following forces :
(i) The weight W at G
(ii) The reaction R at A M S GB
h
(iii) The reaction S at C perpendicular to AB. K Cθ
Since the rod is in equilibrium. So, the three force are concurrent at O. θ
In ∆ACK, we have, sin = h θ R
AC O
θ
In ∆ACO, we have, sin θ = AO W
a N
In ∆AGO, we have sin θ = AO ; ∴ sin 3 θ = h . AC . AO = h
a AC AO a a
A beam whose centre of gravity divides it into two portions a and b, is placed inside a smooth horizontal sphere. If θ be its
inclination to the horizon in the position of equilibrium and 2α be the angle subtended by the beam at the centre of the
sphere, then [Roorkee 1994]
(a) tanθ = (b − a)(b + a) tanα (b) tanθ = (b − a) tanα (c) tanθ = (b + a) tanα (d) tanθ = (b 1 + a) tanα
(b + a) (b − a) − a)(b
Applying m –n theorem in ∆ABC, we get R
(AG + GB)cot ∠OGB = GB cot ∠OAB − AG cot ∠OBG S0
⇒ (a + b) cot(90° −θ ) = b cot π − α − a cot π − α αα B
2 2 θb
b − a tanα A a G M
a + b θ N
⇒ (a + b) tanθ = b tanα − a tanα tanθ
⇒ =
W
Friction. F R
Friction is a retarding force which prevent one body from sliding on another. P
It is, therefore a reaction. W
When two bodies are in contact with each other, then the property of
roughness of the bodies by virtue of which a force is exerted between them to
resist the motion of one body upon the other is called friction and the force
exerted is called force of friction.
(1) Friction is a self adjusting force : Let a horizontal force P pull a heavy body of weight W resting on a
smooth horizontal table. It will be noticed that up to a certain value of P, the body does not move. The reaction R of
the table and the weight W of the body do not have any effect on the horizontal pull as they are vertical. It is the
force of friction F, acting in the horizontal direction, which balances P and prevents the body from moving.
As P is increased, F also increases so as to balance P. Thus F increases with P. A stage comes when P just
begins to move the body. At this stage F reaches its maximum value and is equal to the value of P at that instant.
After that, if P is increased further, F does not increase any more and body begins to move.
This shows that friction is self adjusting, i.e. amount of friction exerted is not constant, but increases gradually
from zero to a certain maximum limit.
(2) Statical friction : When one body tends to slide over the surface of another body and is not on the verge
of motion then the friction called into play is called statical friction.
(3) Limiting friction : When one body is on the verge of sliding over the surface of another body then the
friction called into play is called limiting friction.
(4) Dynamical friction : When one body is actually sliding over the surface of another body the friction
called into play is called dynamical friction.
(5) Laws of limiting friction/statical friction/Dynamical friction :
(i) Limiting friction acts in the direction opposite to that in which the body is about to move.
(ii) The magnitude of the limiting friction between two bodies bears a constant ratio depends only on the
nature of the materials of which these bodies are made.
(iii) Limiting friction is independent of the shape and the area of the surfaces in contact, so long as the normal
reaction between them is same, if the normal reaction is constant.
(iv) Limiting friction fs is directly proportional to the normal reaction R, i.e. fs ∝ R
fs = µs .R; µs = fs / R , where µs is a constant which is called coefficient of statical friction.
In case of dynamic friction, µk = fk/R, where µk is the coefficient of dynamic friction.
(6) Angle of friction : The angle which the resultant force makes with the direction of the normal reaction is
called the angle of friction and it is generally denoted by λ. R S
Thus λ is the limiting value of α, when the force of friction F attains its
maximum value.
∴ tan λ = Maximum force of friction λ
Normal reaction
Since R and µ R are the components of S, we have, S cos λ = R, S sinλ = µR. F=µR
Hence by squaring and adding, we get S = R 1 + µ2 and on dividing
them, we get tan λ = µ. Hence we see that the coefficient of friction is equal to the tangent of the angle of friction.
Coefficient of Friction.
When one body is in limiting equilibrium in contact with another body, the constant ratio which the limiting
force of friction bears to normal reaction at their point of contact, is called the coefficient of friction and it is
generally denoted by µ.
Thus, µ is the ratio of the limiting friction and normal reaction.
Hence, µ = tan λ = Maximum force of friction
Normal reaction
⇒ µ = F ⇒ F = µR, where F is the limiting friction and R is the normal reaction.
R
Note : The value of µ depends on the substance of which the bodies are made and so it differs from one
body to the other. Also, the value of µ always lies between 0 and 1. Its value is zero for a perfectly
smooth body.
Cone of friction : A cone whose vertex is at the point of contact of two rough bodies and whose
axis lies along the common normal and whose semi-vertical angle is equal to the angle of friction is
called cone of friction.
Limiting equilibrium on an Inclined Plane
Let a body of weight W be on the point of sliding down a plane which is inclined at an angle α to the horizon.
Let R be the normal reaction and µ R be the limiting friction acting up the plane.
Thus, the body is in limiting equilibrium under the action of three forces : R µR
R, µ R and W. A
α
Resolving the forces along and perpendicular to the plane, we have
µR = W sinα and R = W cosα W
⇒ µR = W sinα ⇒ µ = tanα ⇒ tan λ = tanα ⇒ α = λ α
R cos α O
Thus, if a body be on the point of sliding down an inclined plane under its own weight, the inclination of the
plane is equal to the angle of the friction.
(1) Least force required to pull a body up an inclined rough plane : Let a body of weight W be at
point A, α be the inclination of rough inclined plane to the horizontal and λ be the P
angle of friction. Let P be the force acting at an angle θ with the plane required Rθ
just to move body up the plane.
A
P = W sin(α + λ) { µ = tan λ} µR α
cos(θ − λ) W
Clearly, the force P is least when cos(θ − λ) is maximum, i.e. when α
cos(θ − λ) = 1 , i.e. θ − λ = 0 or θ = λ . The least value of P is W sin(α + λ) O
(2) Least force required to pull a body down an inclined plane : Let a body of weight W be at the point
A, α be the inclination of rough inclined plane to the horizontal and λ be the angle R µR
of friction. Let P be the force acting an angle θ with the plane, required just to
move the body up the plane. Pθ A
α
P = W sin(λ − α) [ µ = tan λ]
cos(θ − λ) W
Clearly, P is least when cos(θ − λ) is maximum, i.e. when θ − λ = 0 or α
O
θ = λ . The least value of P is W sin(λ − α) .
Note : If α = λ , then the body is in limiting equilibrium and is just on the point of moving downwards.
If α < λ , then the least force required to move the body down the plane is W sin(λ − α).
If α = λ,α > λ or α < λ , then the least force required to move the body up the plane is W sin(α + λ) .
If α > λ, then the body will move down the plane under the action of its weight and normal reaction.
Important Tips
• Least force on the horizontal plane : Least force required to move the body with weight W on the rough horizontal plane is W sin λ .
Example: 20 A force of 35 Kg is required to pull a block of wood weighing 140 Kg on a rough horizontal surface. The coefficient of
Solution: (d)
friction is [BIT Ranchi 1995]
(a) 1 (b) 0 (c) 4 (d) 1
4
In the position of limiting equilibrium, we have µR = 35 and R = 140 ⇒ µ = 35 = 1
140 4
R
µR 35 kg
140 kg
Example: 21 A uniform ladder rests in limiting equilibrium, its lower end on a rough horizontal plane and its upper end against a smooth
Solution: (b)
vertical wall. If θ is the angle of inclination of the ladder to the vertical wall and µ is the coefficient of friction, then tanθ is
Example: 22
Solution: (b) equal to [MNR 1991; UPSEAT 2000]
(a) µ (b) 2µ (c) 3µ (d) µ + 1
2
Resolving the forces horizontally and vertically, we get
S = µR and R = W SB
θ
⇒ S = µW .....(i)
G
Taking moments about A, we get RW
–W.AG sinθ + S.AB cos θ = 0
⇒ W.AG sinθ = S.AB cosθ ⇒ W. AB sinθ = S.AB cosθ AG = AB
2 2
A µR 0
W
⇒ 2 .AB sinθ = µW.AB cosθ [from (i)]
⇒ tanθ = 2µ .
A body of 6 Kg. rests in limiting equilibrium on an inclined plane whose slope is 30°. If the plane is raised to slope of 60°,
the force in Kg. weight along the plane required to support it is [MNR 1987]
(a) 3 (b) 2 3 R
(d) 3 3 µR
(c) 3
30° 6 kg
In case (i),
R = 6 cos 30° , µR = 6 sin 30° .
∴ µ = tan 30° = 1
3
In case (ii), µS
S = 6 cos 60° RP
P + µS = 6 sin 60°
6 kg
∴P + 1 (6 cos 60°) = 6 sin 60° = 6 3 =3 3. 60°
3 2
∴P = 3 3− 1 6× 1 = 3 3− 3 =3 3− 3 =2 3.
3 2 3
Example: 23 The coefficient of friction between the floor and a box weighing 1 ton if a minimum force of 600 Kgf is required to start the
Solution: (b)
box moving is [SCRA 1995]
Example: 24
Solution: (c) (a) 1 (b) 3 (c) 1 (d) 1
4 4 2
Resolving horizontally and vertically R
P cos θ = µR; P sinθ + R = W P
∴ P cosθ = µ[W − P sinθ ] µR θ
or P[cosθ + µ sinθ ] = µW
or P = µW = µW cos λ = W sin λ
cos(θ − λ) cos(θ − λ)
cosθ + sin λ . sinθ W
cos λ
Now P is minimum when cos(θ − λ) is maximum, i.e. when cos(θ − λ) = 1
∴Min P = W sin λ
But W = 1 ton wt. = 1000Kg. and P = 600 kg
∴ sin λ = P = 600 = 3 ; ∴ tan λ = 3 ,∴ µ = 3
W 1000 5 4 4
A block of mass 2 Kg. slides down a rough inclined plane starting from rest at the top. If the inclination of the plane to the
horizontal is θ with tanθ = 4 , the coefficient of friction is 0.3 and the acceleration due to gravity is g= 9.8. The velocity
5
of the block when it reaches the bottom is
(a) 6.3 (b) 5.2 (c) 7 (d) 8.1
Let P be the position of the man at any time. A R
Clearly, R = 2g cosθ µR
Let f be acceleration down the plane. P
Equation of motion is 2 f = 2g sinθ − µR 2g sin θ
2 f = 2g sinθ − µ(2g cosθ ) θ θ
2g cos θ 2g C
2 f = 2g(sinθ − µ cosθ )
B
Here, tanθ = 4 , sinθ = 4 , cosθ = 5
5 41 41
Now, 2 f = 2g 4 − 3 . 5
41 10 41
2f = 2g 4 − 3 = 2g . 5 = 5g , ∴ f = 5g 41 4
41 2 41 2 41 2 41
θ
Let v be the velocity at C. 5
Then, v2 = u2 + 2 fS = 0 + 2 5g AC
2 41
v2 = 5g . 41 we can take AC = 41, since tan θ = 4
41 5
v2 = 5g = 5 × 9.8 = 49.0 , i.e., v2 = 7m / sec
Example: 25 A circular cylinder of radius r and height h rests on a rough horizontal plane with one of its flat ends on the plane. A
Solution: (b)
gradually increasing horizontal force is applied through the centre of the upper end. If the coefficient of friction is µ. The
cylinder will topple before sliding of [UPSEAT 1994]
(a) r < µh (b) r ≥ µh (c) r ≥ 2µh (d) r = 2µh
Let base of cylinder is AB.
∴ BC = r R
Let force P is applied at O. E
Let reaction of plane is R and force of friction is µR. Let weight of cylinder is W. ha DP
In equilibrium condition, µR
O
R=W .....(i) and P = µR .....(ii) W
From (i) and (ii), we have P = µW
Taking moment about the point O,
We have W × BC − P × OC =0 ⇒ P = W × BC = W×r
OC h
If W×r ≥ µW or r ≥ µh
h
The cylinder will be topple before sliding.
Centre of Gravity
The centre of gravity of a body or a system of particles rigidly connected together, is that point through which
the line of action of the weight of the body always passes in whatever position the body is placed and this point is
called centroid. A body can have one and only one centre of gravity.
If w1,w2 ,..........., wn are the weights of the particles placed at the points
A1(x1, y1), A2(x2, y2),.........., An(xn, yn) respectively, then the centre of gravity G(x, y) is given by
w1 x1∑∑ ∑∑x = ,y = w1y1 .
w1 w1
(1) Centre of gravity of a number of bodies of different shape :
(i) C.G. of a uniform rod : The C.G. of a uniform rod lies at its mid-point.
(ii) C.G. of a uniform parallelogram : The C.G. of a uniform parallelogram is the point of inter-section of
the diagonals.
(iii) C.G. of a uniform triangular lamina : The C.G. of a triangle lies on a median at a distance from the
base equal to one third of the medians.
(2) Some Important points to remember :
(i) The C.G. of a uniform tetrahedron lies on the line joining a vertex to the C.G. of the opposite face, dividing
this line in the ratio 3 : 1.
(ii) The C.G. of a right circular solid cone lies at a distance h/4 from the base on the axis and divides it in the
ratio 3 : 1.
(iii) The C.G. of the curved surface of a right circular hollow cone lies at a distance h/3 from the base on the
axis and divides it in the ratio 2 : 1
(iv) The C.G. of a hemispherical shell at a distance a/2 from the centre on the symmetrical radius.
(v) The C.G. of a solid hemisphere lies on the central radius at a distance 3a/8 from the centre where a is the radius.
(vi) The C.G. of a circular arc subtending an angle 2α at the centre is at a distance a sinα from the centre on
α
the symmetrical radius, a being the radius, and α in radians.
(vii) The C.G. of a sector of a circle subtending an angle 2α at the centre is at a distance 2a sin α from the
3 α
centre on the symmetrical radius, a being the radius and α in radians.
(viii) The C.G. of the semi circular arc lies on the central radius at a distance of 2a from the boundry
π
diameter, where a is the radius of the arc.
Important Tips
• Let there be a body of weight w and x be its C.G. If a portion of weight w1 is removed from it and x1 be the C.G. of the removed portion.
Then, the C.G. of the remaining portion is given by x2 = wx − w1 x1
w − w1
• Let x be the C.G. of a body of weight w. If x1, x2, x3 are the C.G. of portions of weights w1, w2, w3 respectively, which are removed from
the body, then the C.G. of the remaining body is given by x4 = wx − w1 x1 − w2 x 2 − w3 x3
w − w1 − w2 − w3
Example: 26 Two uniform solid spheres composed of the same material and having their radii 6 cm and 3 cm respectively are firmly
Solution: (a)
united. The distance of the centre of gravity of the whole body from the centre of the larger sphere is [MNR 1980]
Example: 27
Solution: (b) (a) 1 cm. (b) 3 cm. (c) 2 cm. (d) 4 cm.
Weights of the spheres are proportional to their volumes.
Let P be the density of the material, then
w1 = Weight of the sphere of radius 6cm = 4 π (6 3 )ρ = 288πρ
3
w2 = Weight of the sphere of radius 3cm = 4 π (3 3 )ρ = 36πρ 6 cm 3 cm
3 O O′
x1 = Distance of the C.G. of the larger sphere from its centre O = 0
x2 = Distance of the C.G. of smallar sphere from O = 9 cm.
x = Distance of the C.G. of the whole body from O
Now x = w1 x1 + w2x2 = 288πρ × 0 + 36πρ × 9
w1 + w2 288πρ + 36πρ
x = 36 × 9 =1
324
A solid right circular cylinder is attached to a hemisphere of equal base. It the C.G. of combined solid is at the centre of the
base, then the ratio of the radius and height of cylinder is
(a) 1 : 2 (b) 2 : 1 (c) 1 : 3 (d) None of these
Let a be the radius of the base of the cylinder and h be the height of the cylinder. Let w1 and w2 be the weight of the
cylinder and hemisphere respectively. These weights act at their centres of gravity G1 and G2 respectively.
Now, w1 = weight of the cylinder = πa 2hρg AB
w2 = weight of the hemisphere = 2 πa 3 ρg
3
O1 G1 O2 G2 E
O1G1 = h and O1G2 = h+ 3a DC
2 8
Since the combined C.G. is at O2. Therefore
O1O2 = w1 × O1G + w2 + O1G2
w1 + w2
(πa 2hρg)× h + 2 πa 3 ρg × h + 3a h2 + 2 a h + 3a 2ah h2 2ah a2
2 3 8 2 3 8 3 2 3+ 4
⇒ h = ⇒h= ⇒ h2 + = +
πa 2hρg 2 3 2
+ 3 πa ρg h + 3 a
⇒ 2h2 = a2 ⇒ a = 2 ⇒a:h= 2 :1
h
Example: 28 On the same base AB and on opposite side of it, isosceles triangles CAB and DAB are described whose altitudes are 12 cm
Solution: (b) and 6 cm respectively. The distance of the centre of gravity of the quadrilateral CADB from AB, is
(a) 0.5 cm (b) 1 cm (c) 1.5 cm (d) 2 cm
Let L be the midpoint of AB. Then CL ⊥ AB and DL ⊥ AB. C
12cm G1
Let G1 and G2 be the centres of gravity of triangular lamina CAB and DAB respectively.
Then, LG1 = 1 CL = 4cm. and LG2 = 1 DL = 2cm.
3 3
The C.G. of the quadrilateral ABCD is at G, the mid point of G1 G2. A 6cm L B
∴ G1G2 = GG1 = 3cm. G2
D
⇒ GL = G1L − GG1 = (4 − 3)cm = 1cm.
Example: 29 ABC is a uniform triangular lamina with centre of gravity at G. If the portion GBC is removed, the centre of gravity of the
Solution: (d)
remaining portion is at G'. Then GG' is equal to [UPSEAT 1994]
(a) 1 AG (b) 1 AG (c) 1 AG (d) 1 AG
3 4 5 6
Since G and G' are the centroids of ∆ABC and GBD respectively. Therefore AG = 2 AD,
3
GD = 1 AD and GG" = 2 GD = 2 1 AD = 2 AD
3 3 3 3 9
Now, AG = 2 AD and GD = 1 AD G′
3 3 G
⇒ Area of ∆GBC = 1 Area of ∆ABC G′′
3
⇒ Weight of triangular lamina GBC = 1 (weight of triangle lamina ABC)
3
Thus, if W is the weight of lamina GBC, then the weight of lamina ABC is 3W.
Now, G' is the C.G. of the remaining portion ABGC.
Therefore,
AG' = 3W(AG) − W(AG")
3W − W
= 1 (3 AG − AG" ) AG" = AG + GG" = 2 AD + 2 AD = 8 AD
2 3 9 9
= 1 3 × 2 AD − 8 AD = 5 AD
2 3 9 9
∴ GG' = AG − AG' = 2 AD − 5 AD = 1 AD = 1 3 AG = 1 AG .
3 9 9 9 2 6
Probability
Introduction.
Numerical study of chances of occurrence of events is dealt in probability theory.
The theory of probability is applied in many diverse fields and the flexibility of the theory provides
approximate tools for so great a variety of needs.
There are two approaches to probability viz. (i) Classical approach and (ii) Axiomatic approach.
In both the approaches we use the term ‘experiment’, which means an operation which can produce some
well-defined outcome(s). There are two types of experiments:
(1) Deterministic experiment : Those experiments which when repeated under identical conditions
produce the same result or outcome are known as deterministic experiments. When experiments in science or
engineering are repeated under identical conditions, we get almost the same result everytime.
(2) Random experiment : If an experiment, when repeated under identical conditions, do not produce the
same outcome every time but the outcome in a trial is one of the several possible outcomes then such an
experiment is known as a probabilistic experiment or a random experiment.
In a random experiment, all the outcomes are known in advance but the exact outcome is unpredictable.
For example, in tossing of a coin, it is known that either a head or a tail will occur but one is not sure if a head
or a tail will be obtained. So it is a random experiment.
Definitions of Various Terms.
(1) Sample space : The set of all possible outcomes of a trial (random experiment) is called its sample space.
It is generally denoted by S and each outcome of the trial is said to be a sample point.
Example : (i) If a dice is thrown once, then its sample space is S = {1, 2, 3, 4, 5, 6}
(ii) If two coins are tossed together then its sample space is S = {HT, TH, HH, TT}.
(2) Event : An event is a subset of a sample space.
(i) Simple event : An event containing only a single sample point is called an elementary or simple event.
Example : In a single toss of coin, the event of getting a head is a simple event.
Here S = {H, T} and E = {H}
(ii) Compound events : Events obtained by combining together two or more elementary events are known
as the compound events or decomposable events.
For example, In a single throw of a pair of dice the event of getting a doublet, is a compound event because
this event occurs if any one of the elementary events (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) occurs.
(iii) Equally likely events : Events are equally likely if there is no reason for an event to occur in preference
to any other event.
Example : If an unbiased die is rolled, then each outcome is equally likely to happen i.e., all elementary
events are equally likely.
(iv) Mutually exclusive or disjoint events : Events are said to be mutually exclusive or disjoint or
incompatible if the occurrence of any one of them prevents the occurrence of all the others.
Example : E = getting an even number, F = getting an odd number, these two events are mutually exclusive,
because, if E occurs we say that the number obtained is even and so it cannot be odd i.e., F does not occur.
A1 and A2 are mutually exclusive events if A1 ∩ A2 = φ .
(v) Mutually non-exclusive events : The events which are not mutually exclusive are known as compatible
events or mutually non exclusive events.
(vi) Independent events : Events are said to be independent if the happening (or non-happening) of one
event is not affected by the happening (or non-happening) of others.
Example : If two dice are thrown together, then getting an even number on first is independent to getting an
odd number on the second.
(vii) Dependent events : Two or more events are said to be dependent if the happening of one event affects
(partially or totally) other event.
Example : Suppose a bag contains 5 white and 4 black balls. Two balls are drawn one by one. Then two
events that the first ball is white and second ball is black are independent if the first ball is replaced before drawing
the second ball. If the first ball is not replaced then these two events will be dependent because second draw will
have only 8 exhaustive cases.
(3) Exhaustive number of cases : The total number of possible outcomes of a random experiment in a
trial is known as the exhaustive number of cases.
Example : In throwing a die the exhaustive number of cases is 6, since any one of the six faces marked with 1,
2, 3, 4, 5, 6 may come uppermost.
(4) Favourable number of cases : The number of cases favourable to an event in a trial is the total number
of elementary events such that the occurrence of any one of them ensures the happening of the event.
Example : In drawing two cards from a pack of 52 cards, the number of cases favourable to drawing 2 queens is 4 C2 .
(5) Mutually exclusive and exhaustive system of events : Let S be the sample space associated with a
random experiment. Let A1, A2, …..An be subsets of S such that
(i) Ai ∩ Aj = φ for i ≠ j and (ii) A1 ∪ A2 ∪ .... ∪ An = S
Then the collection of events A1, A2,....., An is said to form a mutually exclusive and exhaustive system of events.
If E1, E2,....., En are elementary events associated with a random experiment, then
(i) Ei ∩ E j = φ for i ≠ j and (ii) E1 ∪ E2 ∪ .... ∪ En = S
So, the collection of elementary events associated with a random experiment always form a system of
mutually exclusive and exhaustive system of events.
In this system, P(A1 ∪ A2....... ∪ An ) = P(A1) + P(A2 ) + ..... + P(An ) = 1 .
Important Tips
• Independent events are always taken from different experiments, while mutually exclusive events are taken from a single experiment.
• Independent events can happen together while mutually exclusive events cannot happen together.
• Independent events are connected by the word “and” but mutually exclusive events are connected by the word “or”.
Example: 1 Two fair dice are tossed. Let A be the event that the first die shows an even number and B be the event that second die
Solution: (d)
shows an odd number. The two events A and B are [IIT 1979]
(a) Mutually exclusive (b) Independent and mutually exclusive
(c) Dependent (d) None of these
They are independent events but not mutually exclusive.
Example: 2 The probabilities of a student getting I, II and III division in an examination are respectively 1 , 3 and 1 . The
Solution: (d) 10 5 4
probability that the student fail in the examination is [MP PET 1997]
(a) 197 (b) 27 (c) 83 (d) None of these
200 200 100
A denote the event getting I; B denote the event getting II;
C denote the event getting III; and D denote the event getting fail.
Obviously, these four events are mutually exclusive and exhaustive, therefore
P(A) + P(B) + P(C) + P(D) = 1 ⇒ P(D) = 1 − 0.95 = 0.05 .
Classical definition of Probability.
If a random experiment results in n mutually exclusive, equally likely and exhaustive outcomes, out of which
m are favourable to the occurrence of an event A, then the probability of occurrence of A is given by
P(A) = m = Number of outcomes favourable to A
n Number of total outcomes
It is obvious that 0 ≤ m ≤ n. If an event A is certain to happen, then m = n, thus P(A) = 1.
If A is impossible to happen, then m = 0 and so P(A) = 0. Hence we conclude that
0 ≤ P(A) ≤ 1.
Further, if A denotes negative of A i.e. event that A doesn’t happen, then for above cases m, n; we shall have
P(A) = n −m = 1 − m = 1 − P(A)
n n
∴ P(A) + P(A) = 1.
Notations : For two events A and B,
(i) A′ or A or AC stands for the non-occurrence or negation of A.
(ii) A ∪ B stands for the occurrence of at least one of A and B.
(iii) A ∩ B stands for the simultaneous occurrence of A and B.
(iv) A′ ∩ B′ stands for the non-occurrence of both A and B.
(v) A ⊆ B stands for “the occurrence of A implies occurrence of B”.
Some important remarks about Coins, Dice , Playing cards and Envelopes.
(1) Coins : A coin has a head side and a tail side. If an experiment consists of more than a coin, then coins
are considered to be distinct if not otherwise stated.
Number of exhaustive cases of tossing n coins simultaneously (or of tossing a coin n times) = 2n.
(2) Dice : A die (cubical) has six faces marked 1, 2, 3, 4, 5, 6. We may have tetrahedral (having four faces 1,
2, 3, 4) or pentagonal (having five faces 1, 2, 3, 4, 5) die. As in the case of coins, if we have more than one die,
then all dice are considered to be distinct if not otherwise stated.
Number of exhaustive cases of throwing n dice simultaneously (or throwing one dice n times) = 6n.
(3) Playing cards : A pack of playing cards usually has 52 cards. There are 4 suits (Spade, Heart, Diamond
and Club) each having 13 cards. There are two colours red (Heart and Diamond) and black (Spade and Club) each
having 26 cards.
In thirteen cards of each suit, there are 3 face cards or coart cards namely king, queen and jack. So there are
in all 12 face cards (4 kings, 4 queens and 4 jacks). Also there are 16 honour cards, 4 of each suit namely ace, king,
queen and jack.
(4) Probability regarding n letters and their envelopes : If n letters corresponding to n envelopes are
placed in the envelopes at random, then
(i) Probability that all letters are in right envelopes = 1 .
n!
(ii) Probability that all letters are not in right envelopes =1− 1 .
n!
(iii) Probability that no letter is in right envelopes = 1 − 1 + 1 − ... + (−1)n 1 .
2! 3! 4! n!
(iv) Probability that exactly r letters are in right envelopes = 1 1 − 1 + 1 − ..... + (−1)n−r (n 1 r)! .
r! 2! 3! 4! −
Example: 3 If (1 + 3p) / 3, (1 − p) / 4 and (1 − 2p) / 2 are the probabilities of three mutually exclusive events, then the set of all values of
Solution: (a)
p is [IIT 1986; AMU 2002; AIEEE 2003]
Example: 4
Solution: (b) (a) 1 ≤ p ≤ 1 (b) 1 < p < 1 (c) 1 ≤ p ≤ 2 (d) 1 < p < 2
Example: 5 3 2 3 2 2 3 2 3
Solution: (b)
Since (1 +3 p) , (1 − p) and 1 − 2p are the probabilities of the three events, we must have
3 4 2
0≤ 1+ 3p ≤ 1, 0 ≤ 1− p ≤1 and 0≤ 1 − 2p ≤1 ⇒ −1 ≤ 3p ≤ 2, − 3 ≤ p≤1 and −1 ≤ 2p ≤ 1
3 4 2
⇒ − 1 ≤ p ≤ 2 , − 3≤ p≤1 and − 1 ≤ p ≤ 1 .
3 3 2 2
Also as 1 + 3p , 1 − p and 1 − 2p are the probabilities of three mutually exclusive events,
3 4 2
0 ≤ 1 + 3p + 1 − p + 1 − 2p ≤1 ⇒ 0 ≤ 4 + 12p + 3 − 3p + 6 − 12p ≤ 12 ⇒ 1 ≤ p ≤ 13
3 4 2 3 3
Thus the required values of p are such that max− 1 , − 3, − 1 , 1 ≤ p ≤ min 2 ,1, 1 , 13 ⇒ 1 ≤ p ≤ 1 .
3 2 3 2 3 3 2
3
The probability that a leap year selected randomly will have 53 Sundays is [MP PET 1991, 93, 95]
(a) 1 (b) 2 (c) 4 (d) 4
7 7 53 49
A leap year contain 366 days i.e. 52 weeks and 2 days, clearly there are 52 Sundays in 52 weeks.
For the remaining two days, we may have any of the two days
(i) Sunday and Monday, (ii) Monday and Tuesday, (iii) Tuesday and Wednesday, (iv) Wednesday and Thursday,
(v) Thursday and Friday, (iv) Friday and Saturday and (vii) Saturday and Sunday.
Now for 53 Sundays, one of the two days must be Sundays, hence required probability = 2 .
7
Three identical dice are rolled. The probability that same number will appear on each of them will be
[SCRA 1991; MP PET 1989; IIT 1984; Rajasthan PET 2000, 02; DCE 2001]
(a) 1 (b) 1 (c) 1 (d) 3
6 36 18 28
If three identical dice are rolled then total number of sample points = 6 × 6 × 6 = 216 .
Favourable events (same number appear on each dice) are
(1, 1, 1) (2, 2, 2) ………(6, 6, 6). ∴ Required probability = 6 = 1 .
216 36
Problems based on Combination and Permutation.
(1) Problems based on combination or selection : To solve such kind of problems, we use n Cr = n! r)! .
r!(n −
Example: 6 Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three
vertices is equilateral, is equal to [IIT 1995; MP PET 2002]
(a) 1 (b) 1 (c) 1 (d) 1
2 5 10 20
Solution: (c) Total number of triangles which can be formed = 6 C3 = 6×5×4 = 20
1× 2×3
Number of equilateral triangles = 2. ∴ Required probability = 2 = 1 .
20 10
Example: 7 Three distinct numbers are selected from 100 natural number. The probability that all the three numbers are divisible by 2
and 3 is [IIT Screening 2004]
(a) 4 (b) 4 (c) 4 (d) 4
25 35 55 1155
Solution: (d) The numbers should be divisible by 6. Thus the number of favourable ways is 16 C3 (as there are 16 numbers in first 100
natural numbers, divisible by 6). Required probability is 16 C3 = 16 ×15 ×14 = 4 .
100 C3 100 × 99 × 98 1155
Example: 8 Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. The chance that the numbers on them
are in A.P., is [Roorkee 1988; DCE 1999]
(a) 10 (b) 9 (c) 9 (d) None of these
133 133 1330
Solution: (a) Total number of ways = 21C3 = 1330 . If common difference of the A.P. is to be 1, then the possible groups are 1, 2, 3; 2,
3, 4; ……19, 20, 21.
If the common difference is 2, then possible groups are 1, 3, 5; 2, 4, 6; ….. 17, 19, 21.
Proceeding in the same way, if the common difference is 10, then the possible group is 1, 10, 21.
Thus if the common difference of the A.P. is to be ≥ 11, obviously there is no favourable case.
Hence, total number of favourable cases = 19 +17 + 15 + …+ 3 + 1 =100
Hence, required probability = 100 = 10 .
1330 133
(2) Problems based on permutation or arrangement : To solve such kind of problems, we use n Pr = (n n! .
− r)!
Example: 9 There are four letters and four addressed envelopes. The chance that all letters are not dispatched in the right envelope is
Solution: (c) [Rajasthan PET 1997; MP PET 1999; DCE 1999]
Example: 10
(a) 19 (b) 21 (c) 23 (d) 1
Solution: (b) 24 23 24 24
Required probability is 1 – P (they go in concerned envelopes) =1− 1 = 23 .
4! 24
The letters of the word ‘ASSASSIN’ are written down at random in a row. The probability that no two S occur together is
[BIT Ranchi 1990; IIT 1983]
(a) 1 (b) 1 (c) 1 (d) None of these
35 14 15
Total ways of arrangements = 8! . •w• x • y • z •
2!.4!
Now ‘S’ can have places at dot’s and in places of w, x, y, z we have to put 2A’s, one I and one N.
Therefore, favourable ways = 5C4 4! . Hence, required probability = 5.4! 2!4! = 1 .
2! 2!8! 14
Odds In favour and Odds against an Event.
As a result of an experiment if “a” of the outcomes are favourable to an event E and “b” of the outcomes are
against it, then we say that odds are a to b in favour of E or odds are b to a against E.
Thus odds in favour of an event E = Number of favourable cases = a = a /(a + b) = P(E) .
Number of unfavourable cases b b /(a + b) P(E)
Similarly, odds against an event E = Number of unfavourable cases = b = P(E) .
Number of favourable cases a P(E)
Important Tips
• If odds in favour of an event are a : b, then the probability of the occurrence of that event is a and the probability of non-occurrence
a+b
of that event is b .
a+b
• If odds against an event are a : b, then the probability of the occurrence of that event is b and the probability of non-occurrence of
a+b
that event is a .
a+b
Example: 11 Two dice are tossed together. The odds in favour of the sum of the numbers on them as 2 are [Rajasthan PET 1987]
(a) 1 : 36 (b) 1 : 35 (c) 35 : 1 (d) None of these
Solution: (b) If two dice are tossed, total number of events = 6 × 6 = 36.
Favourable event is (1, 1). Number of favourable events = 1
∴ odds in favour = 1 = 1 .
36 − 1 35
Example: 12 A party of 23 persons take their seats at a round table. The odds against two persons sitting together are [Rajasthan PET 1999]
(a) 10 : 1 (b) 1 : 11 (c) 9 : 10 (d) None of these
Solution: (a) P = (21)! 2! = 1 = 1 . ∴ odd against = 10 : 1.
(22)! 11 1 + 10
Addition Theorems on Probability.
Notations : (i) P(A + B) or P(A ∪ B) = Probability of happening of A or B
= Probability of happening of the events A or B or both
= Probability of occurrence of at least one event A or B
(ii) P(AB) or P(A∩B) = Probability of happening of events A and B together.
(1) When events are not mutually exclusive : If A and B are two events which are not mutually exclusive,
then P(A ∪ B) = P(A) + P(B) − P(A ∩ B) or P(A + B) = P(A) + P(B) − P(AB) .
For any three events A, B, C
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(C ∩ A) + P(A ∩ B ∩ C)
or P(A + B + C) = P(A) + P(B) + P(C) − P(AB) − P(BC) − P(CA) + P(ABC).
(2) When events are mutually exclusive : If A and B are mutually exclusive events, then
n(A ∩ B) = 0 ⇒ P(A ∩ B) = 0
∴ P(A ∪ B) = P(A) + P(B).
For any three events A, B, C which are mutually exclusive,
P(A ∩ B) = P(B ∩ C) = P(C ∩ A) = P(A ∩ B ∩ C) = 0 ∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) .
The probability of happening of any one of several mutually exclusive events is equal to the sum of their
probabilities, i.e. if A1, A2.....An are mutually exclusive events, then
∑ ∑P(A1 + A2 + ... + An ) = P(A1) + P(A2 ) + ..... + P(An ) i.e. P( Ai ) = P(Ai ) .
(3) When events are independent : If A and B are independent events, then P(A ∩ B) = P(A).P(B)
∴ P(A ∪ B) = P(A) + P(B) − P(A).P(B) .
(4) Some other theorems
(i) Let A and B be two events associated with a random experiment, then
(a) P(A ∩ B) = P(B) − P(A ∩ B) (b) P(A ∩ B) = P(A) − P(A ∩ B)
If B ⊂ A, then
(a) P(A ∩ B) = P(A) − P(B) (b) P(B) ≤ P(A)
Similarly if A ⊂ B, then
(a) (A ∩ B) = P(B) − P(A) (b) P(A) ≤ P(B).
Note : Probability of occurrence of neither A nor B is P(A ∩ B) = P(A ∪ B) = 1 − P(A ∪ B) .
(ii) Generalization of the addition theorem : If A1, A2,....., An are n events associated with a random
P n n n n
i=1 i=1
P(Ai ) − P(Ai ∩ Aj ) + P(Ai
i, j=1 i, j,k=1
∑ ∑ ∑experiment, then Ai = ∩ Aj ∩ Ak ) + ... + (−1)n−1 P(A1 ∩ A2 ∩ ..... ∩ An ).
i≠ j i≠ j≠k
∑If all the events n n
Ai (i = 1, 2..., n) are mutually exclusive, then P i=1 Ai = i=1 P(Ai )
i.e. P(A1 ∪ A2 ∪ .... ∪ An ) = P(A1) + P(A2 ) + .... + P(An ) .
(iii) Booley’s inequality : If A1, A2,....An are n events associated with a random experiment, then
∑(a)Pn Ai ≥ n P(Ai ) − (n − 1) ∑(b)PnAi ≤ n P(Ai )
i=1 i=1 i=1 i=1
These results can be easily established by using the Principle of Mathematical Induction.
Important Tips
Let A, B, and C are three arbitrary events. Then
Verbal description of event Equivalent Set Theoretic Notation
(i) Only A occurs
(i) A ∩ B ∩ C
(ii) Both A and B, but not C occur
(ii) A ∩ B ∩ C
(iii) All the three events occur (iii) A ∩ B ∩ C
(iv) At least one occurs (iv) A ∪ B ∪ C
(v) At least two occur (v) (A ∩ B) ∪ (B ∩ C) ∪ (A ∩ C)
(vi) One and no more occurs (vi) (A ∩ B ∩ C)∪ (A ∩ B ∩ C) ∪ (A ∩ B ∩ C)
(vii) Exactly two of A, B and C occur (vii) (A ∩ B ∩ C)∪ (A ∩ B ∩ C) ∪ (A ∩ B ∩ C)
(viii) None occurs (viii) A ∩ B ∩ C = A ∪ B ∪ C
(ix) Not more than two occur (ix) (A∩ B) ∪ (B ∩ C) ∪ (A ∩ C) − (A ∩ B ∩ C)
(x) Exactly one of A and B occurs (x) (A ∩ B) ∪ (A ∩ B)
Example: 13 A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, what
Solution: (c)
is the probability that it is rusted or is a nail [MP PET 1992, 2000]
Example: 14
Solution: (a) (a) 3/16 (b) 5/16 (c) 11/16 (d) 14/16
Example: 15 Let A be the event that the item chosen is rusted and B be the event that the item chosen is a nail.
Solution: (b)
Example: 16 ∴ P(A) = 8 , P(B) = 6 and P(A ∩ B) = 3 / 16
Solution: (c) 16 16
Example: 17
Solution: (a) Required probability = P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 8 + 6 − 3 = 11 .
16 16 16 16
Example: 18
The probability that a man will be alive in 20 years is 3 and the probability that his wife will be alive in 20 years is 2 .
5 3
Then the probability that at least one will be alive in 20 years is [Bihar CEE 1994]
(a) 13 (b) 7 (c) 4 (d) None of these
15 15 15
Let A be the event that the husband will be alive 20 years. B be the event that the wife will be alive 20 years. Clearly A and
B are independent events. ∴ P(A ∩ B) = P(A) P(B) .
Given P(A) = 3 , P(B) = 2 .
5 3
The probability that at least one of them will be alive 20 years is
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = P(A) + P(B) − P(A). P(B) = 3 + 2 − 3 ⋅ 2 = 9 + 10 −6 = 13 .
5 3 5 3 15 15
Let A and B be two events such that P(A) = 0.3 and P(A ∪ B) = 0.8 . If A and B are independent events, then P(B) =
[IIT 1990; UPSEAT 2001, 02]
(a) 5 (b) 5 (c) 3 (d) 2
6 7 5 5
Here P(A ∪ B) = 0.8 , P(A) = 0.3 and A and B are independent events.
Let P(B) = x . ∴ P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ⇒ P(A ∪ B) = P(A) + P(B) − P(A).P(B)
⇒ 0.8 = 0.3 + x − 0.3x ⇒ x = 5 .
7
A card is chosen randomly from a pack of playing cards. The probability that it is a black king or queen of heart or jack is
[Rajasthan PET 1998]
(a) 1/52 (b) 6/52 (c) 7/52 (d) None of these
Let A, B, C are the events of choosing a black king, a queen of heart and a jack respectively.
∴ P(A) = 2 , P(B) = 1 , P(C) = 4
52 52 52
These are mutually exclusive events, ∴ P(A ∪ B ∪ C) = 2 1 4 = 7 .
52 + 52 + 52 52
If A and B are events such that P(A ∪ B) = 3 / 4, P(A ∩ B) = 1 / 4, P(A) = 2 / 3 , then P(A ∩ B) is [AIEEE 2002]
(a) 5/12 (b) 3/8 (c) 5/8 (d) 1/4
P(A ∪ B) = 3 , P(A ∩ B) = 1 , P(A) = 2 ⇒ P(A) = 1 .
4 4 3 3
∴ P(A ∩ B) = P(A) + P(B) − P(A ∪ B) ⇒ 1 = 1 + P(B) − 3 ⇒ P(B) = 2 .
4 3 4 3
P(A ∩ B) = P(B) − P(A ∩ B) = 2 − 1 = 8−3 = 5 .
3 4 12 12
The probability that A speaks truth is 4 , while this probability for B is 3 . The probability that they contradict each other
5 4
when asked to speak on a fact is [AIEEE 2004]
(a) 4 (b) 1 (c) 7 (d) 3
5 5 20 20
Solution: (c) Let E be the event that B speaks truth and F be the event that A speaks truth.
Example: 19
Now P(E) = 75 = 3 and P(F) = 80 = 4 .
Solution: (c) 100 4 100 5
Example: 20
Solution: (b) ∴ P (A and B contradict each other)
Example: 21
Solution: (a) = P [(B tells truth and A tells lie) or (B tells lie and A tells truth)]
= P[(E ∩ F) ∪ (E ∩ F)] = P(E).P(F) + P(E). P(F) = 3 × 1 + 1 × 4 = 7 .
4 5 4 5 20
A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The
probabilities of the student passing in tests I, II, III are p, q and 1 respectively. If the probability that the student is
2
successful is 1 , then [IIT 1986]
2
(a) p = 1, q = 0 (b) p = 2 , q = 1
3 2
(c) There are infinitely many values of p and q (d) All of the above
Let A, B and C be the events that the student is successful in test I, II and III respectively, then P (the student is successful)
= P[(A ∩ B ∩ C′) ∪ (A ∩ B′ ∩ C) ∪ (A ∩ B ∩ C)] = P(A ∩ B ∩ C′) + P(A ∩ B′ ∩ C) + P(A ∩ B ∩ C)
= P(A).P(B).P(C′) + P(A). P(B′) . P(C) + P(A). P(B). P(C) [∵ A, B, C are independent]
= pq1 − 1 + p(1 − q) 1 + pq 1 = 1 p(1 + q) ⇒ 1 = 1 p(1 + q) ⇒ p(1 + q) = 1 .
2 2 2 2 2 2
This equation has infinitely many values of p and q.
A man and his wife appear for an interview for two posts. The probability of the husband’s selection is 1 and that of
7
wife’s selection is 1 . What is the probability that only one of them will be selected. [AISSE 1987; DSSE 1979, 81, 84]
5
(a) 1 (b) 2 (c) 3 (d) None of these
7 7 7
The probability of husband is not selected = 1− 1 = 6 ; The probability that wife is not selected = 1− 1 = 4
7 7 5 5
The probability that only husband is selected = 1 × 4 = 4 ; The probability that only wife is selected = 1 × 6 = 6
7 5 35 5 7 35
Hence, required probability = 6 + 4 = 10 = 2 .
35 35 35 7
If P(B) = 3 , P(A ∩ B ∩ C) = 1 and P(A ∩ B ∩ C) = 1 , then P(B ∩ C) is [IIT Screening 2003]
4 3 3
(a) 1/12 (b) 1/6 (c) 1/15 (d) 1/9
From Venn diagram, we can see that A
P(B ∩ C) = P(B) − P(A ∩ B ∩ C) − P(A ∩ B ∩ C) C
A– ∩ B ∩ C–
= 3 − 1 − 1 = 1 . A ∩ B ∩ C–
4 3 3 12
B
Example: 22 A purse contains 4 copper coins and 3 silver coins, the second purse contains 6 copper coins and 2 silver coins. If a coin is
Solution: (c)
drawn out of any purse, then the probability that it is a copper coin is [Ranchi BIT 1991; MNR 1984; UPSEAT 2000]
(a) 4/7 (b) 3/4 (c) 37/56 (d) None of these
Required probability = 14 + 16 = 37 .
2⋅7 2⋅8 56
Example: 23 The probability of happening an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then
Solution: (b)
the probability of happening neither A nor B is [IIT 1980; DCE 2000]
(a) 0.6 (b) 0.2 (c) 0.21 (d) None of these
P(A ∩ B) = P(A ∪ B) = 1 − P(A ∪ B)
Since A and B are mutually exclusive, so P(A ∪ B) = P(A) + P(B)
Hence, required probability = 1 − (0.5 + 0.3) = 0.2 .
Conditional Probability.
Let A and B be two events associated with a random experiment. Then, the probability of occurrence of A
under the condition that B has already occurred and P(B) ≠ 0, is called the conditional probability and it is denoted
by P(A/B).
Thus, P(A/B) = Probability of occurrence of A, given that B has already happened.
= P(A ∩ B) = n(A ∩ B) .
P(B) n(B)
Similarly, P(B/A) = Probability of occurrence of B, given that A has already happened.
= P(A ∩ B) = n(A ∩ B) .
P(A) n(A)
Note : Sometimes, P(A/B) is also used to denote the probability of occurrence of A when B occurs.
Similarly, P(B/A) is used to denote the probability of occurrence of B when A occurs.
(1) Multiplication theorems on probability
(i) If A and B are two events associated with a random experiment, then P(A ∩ B) = P(A). P(B / A) , if P(A) ≠ 0
or P(A ∩ B) = P(B). P(A / B), if P(B) ≠ 0.
(ii) Extension of multiplication theorem : If A1, A2,...., An are n events related to a random experiment,
then P(A1 ∩ A2 ∩ A3 ∩ .... ∩ An ) = P(A1)P(A2 / A1)P(A3 / A1 ∩ A2 )....P(An / A1 ∩ A2 ∩ ... ∩ An−1),
where P(Ai / A1 ∩ A2 ∩ ... ∩ Ai−1) represents the conditional probability of the event Ai , given that the events
A1, A2 ,....., Ai−1 have already happened.
(iii) Multiplication theorems for independent events : If A and B are independent events associated
with a random experiment, then P(A ∩ B) = P(A). P(B) i.e., the probability of simultaneous occurrence of two
independent events is equal to the product of their probabilities.
By multiplication theorem, we have P(A ∩ B) = P(A). P(B / A) .
Since A and B are independent events, therefore P(B / A) = P(B) . Hence, P(A ∩ B) = P(A). P(B).
(iv) Extension of multiplication theorem for independent events : If A1, A2,...., An are independent
events associated with a random experiment, then P(A1 ∩ A2 ∩ A3 ∩ ... ∩ An ) = P(A1)P(A2 )...P(An ) .
By multiplication theorem, we have
P(A1 ∩ A2 ∩ A3 ∩ ... ∩ An ) = P(A1)P(A2 / A1)P(A3 / A1 ∩ A2 )...P(An / A1 ∩ A2 ∩ ... ∩ An−1)
Since A1, A2,...., An−1, An are independent events, therefore
P(A2 / A1) = P(A2 ), P(A3 / A1 ∩ A2 ) = P(A3 ),...., P(An / A1 ∩ A2 ∩ ... ∩ An−1) = P(An )
Hence, P(A1 ∩ A2 ∩ ... ∩ An) = P(A1)P(A2)....P(An) .
(2) Probability of at least one of the n independent events : If p1, p2, p3 ,........, pn be the probabilities
of happening of n independent events A1, A2, A3 ,........, An respectively, then
(i) Probability of happening none of them
= P(A1 ∩ A2 ∩ A3...... ∩ An) = P(A1).P(A2).P(A3 ).....P(An) = (1 − p1)(1 − p2)(1 − p3 )....(1 − pn) .
(ii) Probability of happening at least one of them
= P(A1 ∪ A2 ∪ A3.... ∪ An) = 1 − P(A1)P(A2)P(A3 )....P(An) = 1 − (1 − p1)(1 − p2)(1 − p3 )...(1 − pn) .
(iii) Probability of happening of first event and not happening of the remaining
= P(A1)P(A2)P(A3).....P(An) = p1(1 − p2)(1 − p3).......(1 − pn)
Example: 24 If 4P(A) = 6, P(B) = 10, P(A ∩ B) = 1 , then P B = [MP PET 2003]
Solution: (a) A
Example: 25
(a) 2 (b) 3 (c) 7 (d) 19
Solution: (a) 5 5 10 60
P B = P(A ∩ B) = (1 / 10) = 2 .
A P(A) (1 / 4) 5
A coin is tossed three times in succession. If E is the event that there are at least two heads and F is the event in which first
throw is a head, then P E = [MP PET 1996]
F
(a) 3 (b) 3 (c) 1 (d) 1
4 8 2 8
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(E) = 4, n(F) = 4 and n(E ∩ F) = 3
∴ P E = P(E ∩ F) = 3/8 = 3 .
F P(F) 4/8 4
Example: 26 Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second an honour
Solution: (c)
card is (before drawing second card first card is not placed again in the pack) [UPSEAQT 1999, 2003]
(a) 1/26 (b) 5/52 (c) 5/221 (d) 4/13
P(E1) = 4 = 1 , P E2 = 15 = 5
52 13 E1 51 17
P(E1 ∩ E2) = P(E1).P E2 = 1 . 5 = 5 .
E1 13 17 221
Example: 27 If A and B are two events such that P(A) ≠ 0 and P(B) ≠ 1, then P A =
B
[IIT 1982; RPET 1995, 2000; DCE 2000; UPSEAT 2001]
(a) 1 − P A (b) 1− P A (c) 1 − P(A ∪ B) (d) P(A)
B B P(B) P(B)
Solution: (c) P A = P(A ∩ B) = P(A ∪ B) = 1− P(A ∪ B) .
Example: 28 B P(B) P(B) P(B)
If A and B are two events such that P(A ∪ B) = P(A ∩ B) , then the true relation is [IIT 1985]
(a) P(A) + P(B) = 0 (b) P(A) + P(B) = P(A) P B
A
(c) P(A) + P(B) = 2P(A) P B (d) None of these
A
Solution: (c) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ⇒ P(A ∩ B) = P(A) + P(B) − P(A ∩ B) { P(A ∩ B) = P(A ∪ B)}
Example: 29
Solution: (a) ⇒ 2P(A ∩ B) = P(A) + P(B) ⇒ 2P( A). P(A ∩ B) = P(A) + P(B) ⇒ 2P(A) P B = P(A) + P(B).
P(A) A
Example: 30
Solution: (a) Let E and F be two independent events. The probability that both E and F happens is 1 and the probability that neither
12
Example: 31
Solution: (d) E nor F happens is 1 , then [IIT 1993]
Example: 32 2
Solution: (b)
(a) P(E) = 1 , P(F) = 1 (b) P(E) = 1 , P(F) = 1 (c) P(E) = 1 , P(F) = 1 (d) None of these
3 4 2 6 6 2
We are given P(E ∩ F) = 1 and P(E ∩ F) = 1
12 2
⇒ P(E).P(F) = 1 …..(i) and P(E ).P(F ) = 1 …..(ii)
12 2
⇒ {1 − P(E)}{(1 − P(F)} = 1 ⇒ 1 + P(E) P(F) − P(E) − P(F) = 1 ⇒ 1 + 1 − [P(E) + P(F)] = 1
2 2 12 2
⇒ P(E) + P(F) = 7 …..(iii)
12
On solving (i) and (iii), we get P(E) = 1 , 1 and P(F) = 1 , 1 .
3 4 4 3
Let p denotes the probability that a man aged x years will die in a year. The probability that out of n men
A1, A2, A3, .....An each aged x, A1 will die in a year and will be the first to die, is [MNR 1987; UPSEAT 2000]
(a) 1 [1 − (1 − p)n ] (b) [1 − (1 − p)n] (c) n 1 1 [1 − (1 − p)n] (d) None of these
n −
Let Ei denotes the event that Ai dies in a year.
Then P(Ei) = p and P(Ei′) = 1 − p for i = 1, 2, ….n
P (none of A1, A2, .....A3 dies in a year) = P(E1′ ∩ E2′ ∩ .....En′ ) = P(E1′) P(E2′ )....P(En′ ) = (1 − p)n ,
because E1, E2, .......En are independent.
Let E denote the event that at least one of A1, A2, .....An dies in a year.
Then P(E) = 1 − P(E1′ ∩ E′2 ∩ ..... ∩ En′ ) = 1 − (1 − p)n
Let F denote the event that A1 is the first to die.
Then P(F / E) = 1 . Also, P(F) = P(E).P(F / E) = 1 [1 − (1 − p)n ] .
n n
A problem of mathematics is given to three students whose chances of solving the problem are 1/3, 1/4 and 1/5
respectively. The probability that the question will be solved is [BIT Ranchi 1991; MP PET 1990]
(a) 2/3 (b) 3/4 (c) 4/5 (d) 3/5
The probabilities of students not solving the problem are 1− 1 = 2 ,1− 1 = 3 and 1− 1 = 4 .
3 3 4 4 5 5
Therefore the probability that the problem is not solved by any one of them = 2 × 3 × 4 = 2 .
3 4 5 5
Hence, the probability that problem is solved = 1− 2 = 3 .
5 5
The probability of happening an event A in one trial is 0.4. The probability that the event A happens at least once in three
independent trials is [IIT 1980; Kurukshetra CEE 1998; DCE 2001]
(a) 0.936 (b) 0.784 (c) 0.904 (d) 0.216
Here P(A) = 0.4 and P(A) = 0.6
Probability that A does not happen at all = (0.6)3 . Thus required probability = 1 − (0.6)3 = 0.784 .
Total Probability and Baye’s rule.
(1) The law of total probability : Let S be the sample space and let E1, E2,.....En be n mutually exclusive
and exhaustive events associated with a random experiment. If A is any event which occurs with E1 or E2 or …. or
En, then P(A) = P(E1) P(A / E1) + P(E2 ) P(A / E2 ) + ... + P(En ) P(A / En ).
(2) Baye’s rule : Let S be a sample space and E1, E2,.....En be n mutually exclusive events such that
n
Ei = S and P(Ei ) > 0 for i = 1, 2, ……, n. We can think of (Ei’s as the causes that lead to the outcome of an
i=1
experiment. The probabilities P(Ei), i = 1, 2, ….., n are called prior probabilities. Suppose the experiment results in
an outcome of event A, where P(A) > 0. We have to find the probability that the observed event A was due to
cause Ei, that is, we seek the conditional probability P(Ei / A). These probabilities are called posterior probabilities,
given by Baye’s rule as P(Ei / A) = P(Ei ).P(A / Ei ) .
n
∑ P(Ek )P(A / Ek )
k =1
Example: 33 In a bolt factory, machines A, B and C manufacture respectively 25%, 35% and 40% of the total bolts. Of their output 5, 4
Solution: (a) and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. Then the probability that the
Example: 34 bolt drawn is defective is
Solution: (c)
(a) 0.0345 (b) 0.345 (c) 3.45 (d) 0.0034
Let E1, E2 , E3 and A be the events defined as follows:
E1 = the bolts is manufactured by machine A; E2 = the bolts is manufactured by machine B; E3 = the bolts is
manufactured by machine C, and A = the bolt is defective.
Then P(E1) = 25 = 1 , P(E2) = 35 , P(E3) = 40 .
100 4 100 100
P(A / E1) = Probability that the bolt drawn is defective given the condition that it is manufactured by machine A = 5/100.
Similarly P(A / E2) = 4 and P(A / E3 ) = 2 .
100 100
Using the law of total probability, we have P(A) = P(E1)P(A / E1) + P(E2)P(A / E2) + P(E3)P(A / E3)
= 25 × 5 + 35 × 4 + 40 × 2 = 0.0345 .
100 100 100 100 100 100
A lot contains 20 articles. The probability that the lot contains 2 defective articles is 0.4 and the probability that the lot
contains exactly 3 defective articles is 0.6. Articles are drawn at random one by one without replacement and tested till all
the defective articles are found. The probability that the testing procedure ends at the twelfth testing is
(a) 9 (b) 19 (c) 99 (d) 19
1900 1000 1900 900
The testing procedure may terminate at the twelfth testing in two mutually exclusive ways.
(I) When lot contains 2 defective articles, (II) When lot contains 3 defective articles.
Consider the following events.
A = Testing procedure ends at the twelfth testing.
A1 = Lot contains 2 defective articles.
A2 = Lot contains 3 defective articles.
Required probability
= P(A) = P(A ∩ A1) ∪ (A ∩ A2) = P(A ∩ A1) + P(A ∩ A2) = P(A1)P(A / A1) + P(A2)P(A / A2)
Now, P(A / A1) = Probability that first 11 draws contain 10 non-defective and one defective and 12th draw contains a
defective article.
= 18 C10 × 2C1 × 1 .
20 C11 9
And P(A / A2) = Probability that first 11 draws contain 9 non defective and 2 defective articles and 12th draw contains a
defective article = 17 C9 ×3 C2 × 1
20 C11 9
Hence, required probability = 0.4 × 18 C10 × 2C1 × 1 + 0.6 × 17 C9 × 3C2 × 1 = 99 .
20 C11 9 20 C11 9 1900
Example: 35 A bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from a
Solution: (d)
randomly chosen bag and is found to be red. The probability that it was drawn from B is [BIT Ranchi 1988; IIT 1976]
Example: 36
Solution: (a) (a) 5 (b) 5 (c) 5 (d) 25
14 16 18 52
Example: 37
Solution: (b) Let E1 be the event that the ball is drawn from bag A, E2 the event that it is drawn from bag B and E that the ball is red.
We have to find P(E2 / E) .
Since both the bags are equally likely to be selected,
we have P(E1 ) = P(E2 ) = 1 . Also P(E / E1) = 3 / 5 and P(E / E2) = 5 / 9 .
2
Hence by Baye’s theorem, we have P(E2 )P(E / E2 ) 1 ⋅ 5 25 .
P(E1 )P(E / E1 ) + P(E2 )P(E / E2 ) 2 9 52
P(E2 / E) = = 1 3 1 5 =
2 5 2 9
⋅ + ⋅
A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is
actually a six, is
(a) 3 (b) 1 (c) 3 (d) None of these
8 5 4
Let E denote the event that a six occurs and A the event that the man reports that it is a ‘6’, we have
P(E) = 1 , P(E′) = 5 , P(A / E) = 3 and P(A / E′) = 1 .
6 6 4 4
13
P(E).P(A / E) 6×4 3
By Baye’s theorem, P(E / A) = P(E).P(A / E) + P(E′)P(A / E′) = 13 51 = 8 .
6×4 + 6×4
A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, then the
probability that the missing cards is black, is
(a) 1 (b) 2 (c) 1 (d) 25 C13
3 3 2 51 C13
Let A1 be the event that the black card is lost, A2 be the event that the red card is lost and let E be the event that first 13
cards examined are red.
Then the required probability = P A1 . We have P(A1) = P(A2) = 1 ; as black and red cards were initially equal in number.
E 2
Also P E = 26 C13 and P E = 25 C13 .
A1 51 C13 A2 51 C13
P A1 P(E / A1)P(A1) 1 ⋅ 26 C13 2
E A1)P(A1) + P(E / A2)P(A2) 2 51 C13 3
The required probability = = = = .
P(E / 1 26 C13 1 25 C13
2 ⋅ 51 C13 + 2 ⋅ 51 C13
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