vedanta Excel In Opt. Mathematics - Book 10
or, 2cosT + cos2T + 1 = 3 - 3cos2T
or, 4cos2T + 2cosT - 2 = 0
or, 2cos2T + cosT - 1 = 0
or, 2cos2T + 2cosT - cosT - 1 = 0
or, 2cosT (cosT + 1) - 1 (cosT + 1) = 0
or, (2cosT - 1) (cosT + 1) = 0
Either, 2cosT - 1 = 0 ........ (i)
cosT + 1 = 0 .......(i)
From equation (i), cosT = 1 ,
2
or, cosT = cos60° or, cos(360° - 60°)
T = 180°
? T = 60° or 300°
T = 300° is rejected
From equation (ii) cosT = -1,
or, cosT = cos180° ?
? T = 60°, 180°, 300°
On checking,
For, T = 60°
3 sinT - cosT = 1
or, 3 . sin60° - cos60° = 1
or, 3 . -2312- 1 = 1
or, 3 2
2
=1
or, 1 = 1 (True)
For T q
or, 3 sin180° - cos180° = 1
or, 3 . 0 - (-1) = 1
? 1 = 1 (True)
For T = 300°
or, 3 sin300° - cos300° = 1
( )or, 3 - 3 - 1 =1
2 2
3 1
or, - 2 - 2 =1
or, -2 = 1 (False) ?
? T = 60°, 180° are required values.
Example 7. Solve cos3T + cosT = cos2T, 0° ≤ T ≤ 180°
Solution: Here, cos3T + cosT = cos2T
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( ) ( )or,
2cos 3T + T . cos 3T - T = cos2T
2 2
or, 2cos2T.cosT - cos2T = 0
or, cos2T (2cosT - 1) = 0
Either, cos2T = 0 ........ (i)
2cosT - 1 = 0 ........ (ii)
From equation (i), cos2T = 0
or, cos2T = cos90°, cos270°
? 2T = 90°, 270°
? T = 45°, 135°
From equation (ii),
2cosT - 1 = 0
or, cosT = 1
2
or, cosT = cos60°
? T = 60°, 300°
? T 45°, 60°, 135° are the required values.
Example 8. Solve 3 + 1 = 4 , (0° ≤ T ≤ 90°)
Solution: sin2T cos2T
Example 9. Here, 3 + 1 = 4
Solution: sin2T cos2T
or, 3 cos2T + sin2T = 4cos2T.sin2T
Dividing both sides by 2, we get,
3 cos2T + 1 sin2T = 2sin2T.cos2T
2 2
or, sin60°. cos2T + cos60°. sin2T = sin4T
or, sin(2T + 60°) = sin4T
or, sin4T = sin(2T + 60°)
or, 4T = 2T + 60°
or, 2T = 60°
? T= 30°
Solve : tanT + tan2T + tanT . tan2T = 1, 0° ≤ T ≤ 360°
Here, tanT + tan2T + tanT.tan2T = 1
or, tanT + tan2T = 1 - tanT. tan2T
or, tanT + tan2T = 1
1 - tanT . tan2T
or, tan 3T= tan45°, tan (180° + 45°), tan(360° + 45°), tan(540° +45°),
tan(720° + 45°), tan(900° + 45°)
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? 3T = 45°, 225°, 405°, 585°, 765°, 945°
? T = 15°, 75°, 135°, 195°, 255°, 315°
Example 10. Solve : tan2T - (1 + 3 ) tanT + 3 = 0, (0° ≤ T ≤ 360°)
Solution: Here, tan2T - (1 + 3 ) tanT + 3 = 0
or, tan2T - tanT - 3 tanT + 3 = 0
or, tanT (tanT - 1) - 3 (tanT - 1) = 0
or, (tanT - 1) (tanT - 3) = 0
Either, tanT - 1 = 0 ....... (i)
tanT - 3 = 0 ........ (ii)
From equation (i), tanT = 1
or, tanT = tan45°, tan225°
From equation (ii), tanT = 3
or, tanT = tan60°, tan240°
? T = 60°, 240°
? T = 45°, 60°, 225°, 240° are the required values.
Example 11. If tanT + tanE = 2 and cosv.cosE = 1 solve for values of D and E
2
Solution: Here, tanD + tanE = 2
or, sinD + sinE =2
cosE cosD
or, sinD.cosE + cosD.sinE = 2cosD. cosE
or, sin(D + E) = 2cosD. cosE ....... (i)
But cosD.cosE = 1 ......... (ii)
2
1
? sin(D + E) = 2 .2
or, sin(D + E) = 1 = sin90°
? D + E = 90°
and E = 90° - D
From (ii) 2cosD.cos(90° - D) = 1
or, 2cosD. sinD = 1
or, sin2D = sin90°
or, 2D = 90° ? D = 45°
Also, D + E = 90°
E = 90° - D = 90° - 45° = 45°
? D = 45°, E = 45°
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Exercise 9.5
Very Short Questions
1. Find the value of T in the following equation, 0° ≤ T ≤ 180°.
(a) 2sinT = 3 (b) 3 tanT = 1 (c) cosT = 1
(d) cotT = 3 (e) secT = 2 2
(f) 2cosT + 1 = 0
(g) 3 tanT - 1 = 0
2. Find the value of T, 0° ≤ T ≤ 90°
(a) tanT = cotT (b) cosT = sinT (c) secT = cosecT
(d) cot2T = tanT (e) sin4T = cos2T (f) cos2T = sin3T
3. Solve the following questions : 0° ≤ T ≤ 180°
(a) sin2T = sinT (b) tanT = sinT (c) 3cotT - tanT = 2
(d) 2 +2=0 (e) tanT + cotT = 2 (f) 1 + cos2T = cosT
cosT
Short Questions
4. Solve, 0° ≤ T ≤ 360° (b) 4sin2T = 3 (c) sec2T = 4
(a) 3tan2T = 1 (e) 3sec2T - 4 = 0
(d) sec2T = 2tan2T
Long Questions
5. Solve for T, 0° ≤ T ≤ 180°
(a) sin4T = cosT - cos7T (b) sinT + sin2T + sin3T = 0
(d) cos2T + cos4T = cos3T
(c) sinT - sin2T = 0 (f) 3cotT - tanT = 2
(e) secT . tanT = 2
6. Solve : 0° ≤ T ≤ 360° (b) 2sin2T - 3sinT + 1 = 0
(a) 2cos2T - 3sinT
(c) 2cos2T - 5cosT + 2 = 0 (d) 2sin2T + sinT - 1 = 0
(e) 4cos2T + 4sinT = 5
(f) 3 - 2sin2T = 3cosT
( )(h) cot2T +
(g) tan2T + (1 - 3 ) tanT = 3 3 1 cotT + 1 = 0
7. Solve : 0° ≤ x ≤ 360° 3
(a) 3 cosx + sinx = 3 (b) sinx + cosx = 2
(d) 3 cosx = 3 - sinx
(c) sinx + cosx = 1, (f) cosx - 3 sinx = 1
(e) 3 sinx + cosx = 2
(g) 2 secx + tanx = 1
8. Solve for x, 0° ≤ x ≤ 180°
(a) 3 + 1 =0 (b) 3 + 1 = 4
sin2x cos2x sin2x cos2x
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9. Solve : 0° ≤ x ≤ 360°
(a) 2 tan3x.cos2x + 1 = tan3x + 2cos2x (b) sin2x . tanx + 1 = sin2x + tanx
10. Find the value of T from.
tanT - 3cotT = 2tan3T, 0° ≤ T ≤ 360°
11. Solve: 0° ≤ x ≤ 90°
tan(x + 100°) = tanx.tan(x + 50°).tan(x – 50°)
1. (a) 60°, 120° (b) 30° (c) 60° (d) 30°
(e) 45° (f) 120° (g) 30°
2. (a) 45° (b) 45° (c) 45° (d) 30°
(e) 15°, 75° (f) 18°, 90°
3. (a) 0°, 60°, 180° (b) 0°, 180° (c) 45°, 108.43° (d) 135°
(e) 45° (f) 60°, 90°
4. (a) 30°, 150°, 210°, 330° (b) 60°, 120°, 240°, 300°
(c) 60°, 120°, 240°, 300° (d) 45°, 135°, 225°, 315°
(e) 30°, 150°, 210°, 330°
5. (a) 0°, 10°, 45°, 50°, 90°, 130°, 135°, 170° (b) 0°, 90°, 120°, 180° (c) 0°, 60°, 180°
(d) 30°, 60°, 90°, 150°, 180° (e) 45°, 135° (f) 45°, 108.43°
6. (a) 30°, 150° (b) 0°, 30°, 90° (c) 60°, 300° (d) 30°, 150°, 270°
(e) 30°, 150° (f) 0°, 60°, 300°, 360°
(g) 60°, 135°, 240°, 315° (h) 120°, 150°, 300°, 330°
7. (a) 0°, 60°, 360° (b) 45° (c) 0°, 90°, 360° (d) 0°, 60°, 360°
(e) 60° (f) 0°, 240° (g) 315°
8. (a) 60°, 150° (b) 20°, 30°, 50°, 80°, 120°, 140°, 170°
9. (a) 15°, 30°, 75°, 150°, 210°, 225°, 315°, 330° (b) 45°, 225°
10. 45°, 60°, 120°, 135°, 225°, 240°, 300°, 315° 11. 30°, 55°
9.6 Height and distance
Trigonometric knowledge can be used to calculate heights of all objects or distance of
inaccessible objects. A
Introduction :
Let AB be a pillar. An observer is at point C which is at a distance BC from 12cm
the foot of the pillar. The observer observes the top of the pillar through
CA. The line CA is called sight line. AB is the height of the pillar. BC is the B 60°
distance between the foot of the pillar and the observer. Let AC = 12m and C
BCA = 60°. Now discuss the following questions.
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(i) What is the height of the pillar ?
(ii) Find the distance between the foot of the pillar and the observer.
Let us define some basic terms used in this topic : Object A
Angle of elevation :
When an object is at a higher level than that of the Sight line
observer, the observer has to have a view of the object Angle of elevation
or the observer has to look up. The angle which the
sight line makes with the horizontal the observers Observe's eye O T
eye is known as the angle of elevation. In the figure
AOB = T is the angle of elevation. O Horizontal Line through B
T observe's eye A
Angle of Depression :
When the object is at a lower level than that of the observer Angle of Depression Sight line
then he has to look below in order to see the object. The C B
angle which the sight line through the observer's eye is
known as the angle of depression. In the above figure
AOB = T is the angle of depression. OA is parallel to
CB, AOB = CBO = T, being the corresponding angles.
How is the angle of elevation or depression measured?
Special instruments like theodolite, sextant, clinometer,
hypsometer etc are used to measure the angle of depression.
Step 1 : Draw the appropriate figure which represents the
given condition.
Step 2 : Identify the given sides and angles in the figure.
Step 3 : Choose the appropriate trigonometric ratio to find the unknown sides or angles.
Step 4 : Carry out the necessary calculations with appropriate units.
Worked Out Examples
Example 1. Find the values of x and y from the given figures:
(a) A (b) S 45° P
xx
D 30° 60° B Q 40m R
20m Cy
Solution: (a) In the given figure from right angle ∆ABC,
tan60° = x
y
or, 3 y = x ? x = 3 y ............ (i)
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Again, from right angled ∆ABD,
tan30° = 20 x y
+
or, 1 = x y
3 20 +
or, 20 + y = 3 x .......... (ii)
Putting the value of x in equation (i), we get
20 + y = 3. 3y
or, 20 + y = 3y
or, 2y = 20 ? y= 20 ? y = 10m
2
Putting the value of y in equation (i)
20 + 10 = 3 x
or, x= 30 = 10 3 ? x = 10 3m
3
(b) In the given figure, SP//QR, corresponding angles.
SPQ = PQR,
Now, from right angled triangle PQR,
tan45° = x or, 1 = x
40 40
? x = 40m
Example 2. A man observes the angle of elevation of the top of a tower to be 30°. On
Solution: walking 200 metre nearer, the elevation is found to be 60°. Find the height of
the tower. Also, find the distance between the first point from the foot of the
tower.
Let CD be the height of the tower, A and B be the first and the second
point respectively. In the figures AB = 200m, suppose BC = ym and
CD = xm. From right angled triangle BCD,
tan60° = CD
BC
or, 3 = x
y
D
or, x = 3 y .......... (i)
Again, from right angled triangle ACD. x
tan30° = CD 30° 60°
AC 200m By
A C
x
or, 1 = 200 + y
3
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or, 200 + y = 3 x ........... (ii)
Putting the value of x in equation (ii)
200 + y = 3. 3 y
or, 200 + y = 3y
or, 200 = 2y
? y = 100m
Putting the value of y in equation (i)
x = 3. 100 = 1.732 × 100 = 173.2 m
? The height of the tower is 173.2m and distance between the tower and
first point is (200 + 100) = 300m
Example 3. From the roof of a house, an observer finds that the angles of depression of
Solution: two places in the same side are 60° and 45°. If the distance between these
places is 40m, find the distance of the places from the foot of the house.
Let PQ be the height of the house and R and S he two points on the same side
of the house.
Then PT//QS, TPR = PRQ = 60°, TPS = PSR = 45°, RS = 40m.
Let QR = y m and PQ = x m. P 60°45° T
Now, from right angled triangle PRQ
tan60° = PQ xm
QR
x
or, 3 = y Q 60° 45° S
y R 40m
or, x = 3 y ............... (i)
Again, from right angled triangle PQS
tan45° = PQ
QS
x
or, 1 = y + 40
or, x = y + 40 ........ (i)
Putting the value of x in equation (ii) from (i), we get
3 y = y + 40
or, y ( 3 - 1) = 40
or, y = 5434.06-14m= 40 - 1 = 40
? y = 1.732 0.732
Putting the value of y in equation (i)
x = 3 × 54.64 = 94.64m
? The height of the house (x) = 94.64m
Distance of the first point from the house (y) = 54.64m
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Distance of the second point from the house (x) = y + 40 = 54.64 + 40
= 94.64m.
Example 4. Two girls are on the opposite sides of a tower which is 200m high. They
Solution: observed that the angles of elevation of the top of the tower are 30° and 45°
respectively. Find the distance between them.
Let A and B be the positions of the two girls on opposite sides of the tower DC.
DC = 200m.
The angles of elevations are
DAC = 30°, DBC = 45°
Let AC = xm and BC = ym
From right angled ∆DAC, D
tan30° = DC 200m
AC
1 200
or, x =3 = x A 30° 45° B
or, 200 3 .......(i) xm C ym
Again, from right angled ∆ACB,
tan45° = CD
CB
200
or, 1= y ? y = 200m
= AB = AC + CB
Distance between the girls
= x + y = 200 3 m + 200m = 546.41m
Example 5. The angle of depression of the top and the foot of a lamp post observed from
Solution: the roof of a 60m high house and found to be 30° and 60°. Find the height of
the lamp post.
Let MN and PQ be the height of the house and lamp post respectively.
Then, MN = 60m, RMP = MPS = 30°, RMQ = MQN = 60°
From right angled, MQN, R 30° M
60°
MN
tan60° = QN
or, 3 = 60 P 30° S
QN
60 60
? QN = 3 = 3 3 = 20 3 m. Q 60°
N
PS = QN = 20 3 (opposite sides of a rectangle)
Again, from right angled triangle MPS,
tan30° = MS
PS
1 MS
or, 3 = 20 3
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MS = 20 3 = 20m
3
Hence, the height of the lamp post is
PQ = SN = MN - MS = 60 - 20 =40m.
Example 6. A boy standing between two pillars of equal height observes the angle of
Solution: elevation of the top of pillars to be 30°. Approaching 15m towards any one
of the pillar, the angle of elevation is 45°. Find the height of the pillars and
the distance between them.
Let PQ and MN be two equal pillars of equal height,
QN the distance between the pillars.
O is mid point of QN. MN = PQ.
MON = 30°, POQ = 30°, RO = 15m.
From right angled triangle PQR, tan45° = PQ
QR
or, QR = PQ
Again, from right angled triangle PQO,
tan30° = PQ P M
QO N
1 PQ
or, 3 = QR + 15
1 PQ Q 45° 30° 30°
3 PQ + 15 R 15m O
or, =
or, 3 PQ = PQ + 15
or, ( 3 - 1 ) PQ = 15
or, PQ = 15 = 15 × 3 +1
3-1 3-1 3 +1
15 ( 3 +1)
= 3-1
= 15 (1.732 + 1) = 7.5 × 2.732
2
= 20.49m.
Again, from the right angled triangle MON,
tan30° = MN or, 1 = 20.49
ON 3 ON
? ON = 35.49m
Distance between two pillars is 2 × ON = 2 × 35.49 = 70.98m.
Hence, the height of the pillar = PQ = MN = 20.49m
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Example 7. A flagstaff stands upon the top of a building. Find the length of the flagstaff
Solution: and building observed from a point at a distance of 20 m from the building
are respectively 60° and 45°.
Let PQ be the flagstaff and QR be the building respectively. P
Let S be a point 20m away from the building, SR = 20m
Now, from right angled triangle SRQ. Q
tan45° = QR 60°
SR 45°
S R
20m
or, 1= QR
20
? QR = 20m
Again, from right angled triangle PRS.
tan60° = PR
SR
or, 3 = PQ + QR
SR
or, 3= PQ + 20
20
or, 20 3 = PQ + 20
? PQ = 20 ( 3 - 1) = 20 (1.732 - 1)
= 20 × 0.732 = 14.64m.
Hence, the height of the flagstaff is 14.64m.
Example 8. PQ and MN are two towers standing on the same horizontal plane. From
Solution: point M the angle of depression of Q is 30° and from N the angle of elevation
of P is 60°. If the shorter tower is MN = 100m, find the height of the tower
PQ.
In given two towers are PQ and MN, MN<PQ. MN = 100m,
RMQ = MQN = 30°
PNQ = 60°. P
From right angled triangle MQN,
tan30° = MN
QN
or, 1 = 100 R 30° M
3 QN Q 30° 60° N
100m
? QN = 100 3 m. 261
RMNQ is a rectangle,
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? RM = QN = 100 3 m.
From right angled triangle PQN,
tan60° = PQ
QN
PQ
or, 3= 100 3
? PQ = 300m
? The height of the longer tower PQ is 300m.
Example 9. From a light house the angles of depression of two ships on the opposite
Solution: sides of the light house were observed to be 30° and 45°. If the height of the
light house is 200m and the line joining the two ships passes through the
foot of the light house, find the distance between the ships.
Let AD be the height of the light house.
EAB = ABD = 45°, FAC = ACD = 30°
From the right angled triangle ABD, tan45° = AD
BD
or, BD = AD
A
? BD = 100m E F
100m
Again, from right angled triangle ADC,
tan30° = AD B 45° D 30° C
DC
or, 1 = 100
3 DC
or, DC = 173.2m.
Hence, the distance between the ships is given by
BC = BD + DC = 100m + 173.2m = 273.2m
Example 10. A ladder of 9 m long reaches a point 9 m below the top of a vertical flagstaff.
From the foot of the ladder the angle of elevation of the flagstaff is 60°. Find
the height of the flagstaff.
Solution: Let MN be the height of the flagstaff and PS be a ladder.
Then MPN = 60°, MS = 9m, PS = 9m M
Let SN = x, PN = y .
From right angled triangle MPN, by using Pythagoras theorem, 9m
PS2 = PN2 + SN2 P 60°y S
or, 81 = y2 + x2 x
? x = 81 - y2 ......... (i)
N
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Again, from right angled triangle MPN,
tan60° = MN
PN
or, 3= 9+x
y
or, 3 y = 9 + x ............ (ii)
By solving equation (i) and (ii), we get
3 y – 9 = 81 – y2
squaring on both sides, we get,
( 3 y – 9)2 = ( 81 – y2)2
or, 3y2 – 18 3y + 81 = 81 – y2
or, 3y2 + y2 = 18 3y
or, 4y2 = 18 3y
or, 4y2 = 18 3
y
18 3
or, y = 4
since y ≠ 0, y = 93 = 7.8
2
and x = 81 – 60.75 = 4.5m
? MN = 9 + 4.5 = 13.5m
Alternative Method
In above diagram, ∆PSM is an isosceles triangle.
? PMS = MPS = 30°
SPN = 60° - 30° = 30°
From, right angled triangle ∆SPN
sinSPN = SN
PS
x
or, sin60° = 9
or, 1 = x
2 9
? x = 4.5 m
Hence, the height of the flagstaff is (9 + 4.5) = 13.5m
Example 11. A vertical column is divided into ratio 1:9 from the bottom by a point on
it. If these two parts make an equal angle 20m from the bottom of the any
column, find the height of the column.
Solution: Let MN be the height of the vertical column and P be the point on it which
divides MN in the ratio of 1:9 from the bottom.
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Let NP = x and PM = 9x.
Let each segment of the column makes an angle T at Q.
PQN = T MQP = T, QN = 20m M
From right angled triangle PQN 9x
tanT = PN = x P
QN 20 x
Again, from right angled triangle MQN,
N
tan2T = MN Q T
QN
T
20m
or, 2tanT = 10x
or, 1- -ta2nx02T 20
1
= x
x2 2
1 - 400
or, x × 400 = x
10 400 - x2 2
or, 400 - x2 = 80
or, x2 = 320
Since height is positive, taking positive square root only,
x = 8 5 m.
The height of the column is 10x = 10 × 8 5 = 80 5 m.
Example 12. The angle of elevation of a bird from a point 80m above the lake is 30°.
From the same point, the angle of depression to its image on the same lake
is found to 60°. Find the height at which the bird is at the moment.
Solution: Let B be the position of the bird and its image is B' in the lake. Then a point A
is 80m above the surface of water, XY the level of water of the lake.
Let BC = x, CB' = y, AD = 80m. CAB = 30°, B'AC = 60°
From right angle triangle ABC, B
x
tan30° = BC C 30°
CA
1 x 60° A
or, 3 = CA XE 80m
DY
or, CA = 3 x ........(i) B'
Again, from right angled triangle ACB'
or, tan60° =3y.xCCBA'[by using (i)]
3=
or, y = 3x ............ (i)
But y = CE + EB' ( BE = B'E)
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B'E = 80 + BC = 80 + x
? y = 80 + 80 + x ....... (iii)
From (ii) and (iii), we get,
160 + x = 3x
or, 2x = 160 ? x = 80m
? The position of the bird from the level of lake is
BE = CE + CB = 80 + 80 = 160m
Example 13. The shadow of a pole on the ground level increases in length by x metres
when the sun's attitude changes from 45° to 30°. Calculate values of x given
that height of the pole is 50m.
Solution: Let MN be the height of the pole, MN = 50m.
Let NP = y and NQ = y + x be the length of shadows when sun's altitudes are
45° and 30° respectively. Increased in length of shadow is PQ = x.
From, right angled triangle MPN, M
50m
tan45° = MN
PN
50
or, 1 = y
or, y = 50m Q 30° 45° N
Again, from right angled triangle MQN. xPy
tan30° = MN
QN
or, 1 = 50
3 x+y
or, 50 3 = x + 50
or, x = 50 ( 3 - 1)
or, x = 50 × (1.732 - 1) = 50 × 0.732 = 36.6m
? x = 36.6m
Exercise 9.6
Very Short Questions
1. Define the following terms with figures:
(a) Angle of elevation (b) Angle of depression
2. Find the value of x in the following diagrams:
PM 30° N
20m
(a) (b)
60m x
Q 60° R P xO
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Short Questions
3. Find the values of x and y in the given diagrams:
P
(a) (b) P
y 45°
S 30° 45° Q 200m
20m R x
Q 60° R
x Sy
(c) P (d) P 30° 60° U
Q 9m 9m 20m 30° R
60° S T x 60° S
x y
y R
Q
Long Questions
4. (a) The angle of elevation of the top of a house from a point on the ground was
observed to be 60°. On walking 60m away from that point, it was found to be 30°.
If the house and points are in the same line of the same plane, find the height of
the house.
(b) The angle of elevation of the top of a tower from a point was observed to be 45°.
On walking 30m away from that point, it was found to be 30°. Find the height of
the house.
5. (a) From a helicopter flying vertically above a straight road, the angles of depressions
of two consecutive kilometer stone on the same side are found to be 45° and 60°.
Find the height of the helocopter.
(b) From the top of a tower of 100m, the angle of depression of two places due to east
of it are respectively 45° and 60°. Find the distance between the two places.
(c) From the top of a tower of 200 m, the angle of depression of two boats which are
in a straight line on the same side of the tower are found to be 30° and 45°. Find
the distance between the boats.
6. (a) From the top of 21 metre high cliff, the angles of depression of the top and bottom
of a towers are observed to be 45° and 60° respectively. Find the height of the
tower.
(b) From the top of a cliff the measure of the angle of depression of the top and bottom
of a building are found to be 30° and 45°. If the height of the cliff is 100 metres.
Find the height of the building.
7. (a) Two poles stand on either side of a road. At the point midway between the two
posts, the angles of inclination(elevation) of their tops are 30° and 60°. Find the
length of the shorter posts if the longer post is of 15m long.
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(b) Two lamp posts are of equal height. A girl standing mid way between them observes
the angle of elevation of either posts to be 45°. After walking 15m towards one of
them, she observes its elevation to be 60°. Find the heights of the posts and the
distance between them.
(c) The posts are 120m apart, and the height of one is double that of other. From the
middle point of the line joining the feet, an observer found the angle of elevation
of their tops to be complementary. Find the height of the longer post.
8. (a) A flagstaff stands upon the top of a building. The angles of elevation of the top
of the building and flagstaff are observed to be 45° and 60° respectively. If the
distance between the observer and the building is 10m, find the length of the
flagstaff.
(b) A flagstaff stands on the top of a tower. The angles subtended by the tower and the
flagstaff at a point 100 metres away from the foot of the tower are found to be 45°
and 15°. Find the length of the flagstaff.
(c) A flagstaff is placed at one corner of a rectangular garden 40m long and 30m wide.
If the angle of elevation of the top of the flagstaff from the opposite corner is 30°.
Find the height of the flagstaff.
9. (a) From the top and bottom of a tower, the angle of depression of the top of the house
and angle of elevation of the house are found to be 60° and 30° respectively. If the
height of the building is 20m. find the height of the tower.
(b) From the roof of a house 30m height, the angle of elevation of the top of a tower is
45° and the angle of depression of its foot is 30°. Find the height of the tower.
(c) A crow is sitting on the top of a tree which is in front of a house. The angle of
elevation and angle of depression of the crow from the bottom and top of the
house are 60° and 30° respectively. If the height of the tree is 21m. Find the height
if the house.
10. (a) A ladder of 18m reaches a point of 18m below the top of vertical flagstaff. From the
foot of the ladder the angle elevation of the flagstaff is 60°. Find the height of the
flagstaff.
(b) A ladder 15m long reaches a point 15 below the top of a vertical flagstaff. From
the foot of the ladder, the angle of elevation of the top of flagstaff is 60°. Find the
height of the flagstaff.
11. (a) A pole is divided by a point in the ratio 1:9 from bottom to top. If the two parts
of the pole subtend equal angles at 20m away from the foot of the pole, find the
height of the pole.
(b) AB is a vertical pole with its foot B on a level of ground. A point C on AB divides
such that AC : CB = 3:2. If the parts AC and CB subtend equal angles at a point on
the ground which is at a distance of 20m from the foot of the pole, find the height
of the pole.
12. (a) A man 1.75m stands at a distance of 8.5m from a lamp post and it is observed that
his shadow is 3.5m long. Find the height of the lamp post.
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(b) A tree of 12m height is broken due to wind such a way that its top touches the
ground and makes an angle of 60° with the ground. At what height from the bottom
is the tree broken by the wind.
13. The angle of elevation of an aeroplane from a point in the ground is 45°. After 15 seconds
of flight the angle if elevation changes to 30°. If the aeroplane is flying horizontally at a
height of 4000m in the same direction, find the speed of aeroplane.
14. The angle of elevation of the top of a tower is 45° from a point 10m above the water
level of a lake. The angle of depression of its image in the lake from the same point is
60°. Find the height of the tower above the water level.
15. A rope dancer was walking on a loose rope tied to the tops of two posts each 8m high.
When the dancer was 2.4m above the ground, it was found that the stretched pieces of
the rope made angles of 30° and 60° with the horizontal line parallel to the ground. Find
the length of the rope.
Project Work
16. Select buildings around your school. Find the angle of elevation by using sextants
or theodolite. Find the height of the buildings taking the distance between the two
places in meter.
17. Write a short note on " Height and Distance" with its application.
2. (a) 30 3 m (b) 20 3 m
3. (a) x = y = 27.32 m (b) x = 200 m, y = 115.47 m
(c) x = 7.79 m, y = 4.5 m (d) x = 20 3, y = 40 m
4. (a) 51.96 m (b) 40.98 m 5.(a) 2365.98 m
(b) 42.26 m (c) 146.4 m 6.(a) 8.87 m (b) 42.27 m
7. (a) 5 m (b) 35.49 m, 70.98 m (c) 84.85 m
8. (a) 7.32 m (b) 73.2 m (c) 28.86 m
9. (a) 80 m (b) 81.96 m (c) 28 m
10. (a) 27 m (b) 22.5 m 11.(a) 178.88 m (b) 22.36 m
12. (a) 6 m (b) 5.57 m
13. 195.2 m/s 14. 37.32 m 15. 17.67 m
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10Vectors
10.0 Review
Discuss the following questions:
a. Classify the following quantities to vectors or scalars with your reasons.
(i) Temperature (ii) Pressure (iii) Force (iv) Length
(v) Area (vi) Acceleration (vii) Displacement (viii) Time
b. Give an example of each of the following vectors:
(i) Column vector (ii) Row vector (iii) unit vector
(iv) Equal vectors (v) Negative vectors (vi) position vectors
(vii) Unequal vectors (viii) Like and unlike vectors.
c. What role do vectors play in our daily life ? Give some examples.
o o oo oo
d. Let a = (2, -3) and b = (1, 2), find a + b and a - b . Express them in the
oo
form of x i + y j .
10.1 Scalar (or Dot) Product of Two Vectors
oo
Let OA = (2, 3), OB = (-3, 2) be two vectors. Plot them in a graph paper. Then,
(a) Find the product of x-components and y-components Y
of the vectors.
(b) Add the products.
(c) What is the result ? BA X
oo X' O
(d) Are OA and OB perpendicular to each other?
Here, 2 × (-3) + 3 × 2 = - 6 + 6 = 0
oo
OA is perpendicular to OB .
There are two types of product of two vectors: Y'
(a) Scalar product or dot product.
(b) Vector product or cross product.
We discuss only about scalar product or dot product of two vectors.
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Scalar Product of Two Vectors
The scalar product of two vectors oa and ob is defined as the
product of the magnitude of two vectors multiplied by the cosine
of the angle T between their directions.
Thus, oa · ob = |oa | |ob | cosT = ab cosT
where, |oa |= a and |ob |= b.
AO = oa and OB = ob
Now, draw perpendicular BM from B to OA.
Now, oa · ob = |oa | |ob |cosT = a b cos T
= (OA) (OB) cos T
= OA (OB cos T)
= (OA) (OM) = (magnitude of oa ) (component of ob in the direction of oa )
So, it is clear that the scalar product of two vectors is equivalent to the product of the
magnitude of one vector with the component of the other vector in the direction of this
vector.
If we write oa · ob , the rotation of oa towards ob is anti-clockwise and the angle T is taken to
be positive.
? oa · ob = ab cosT
If we write ob ;oa , the rotation of ob towards oa is clockwise and the angle T is taken to be
negative.
? ob · oa = b a cos(–T) = ba cosT
Hence, oa · ob = ob · oa
Note :
Let oa = x1oi + y1oj and ob = x2oi + y2oj be two vectors in terms of the components. Then
we know that oi and oj are unit vectors along X-axis and Y-axis respectively.
oi . oj = |oi ||oj |cos 90° = 0, oi . oi = |oi | |oi | cos0° = 1
Now, oa . ob = (x1 oi + y1) . (x2oi + y2oj ) = x1oi .(x2oi + y2oj ) + y1 oj . (x2oi + y2 oj )
= x1x2 oi . oi + x1y2 oi . oj + y1 x2 oj . oi + y1y2 oj . oj = x1x2 + y1y2
? oa . ob = x1x2 + y1y2
Hence, the scalar product of two vectors is equal to the sum of the product of their
corresponding components.
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Two Important Results of Scalar Product of Two Vectors.
(a) If oa and ob are perpendicular to each other, i.e., T = 90° B A
Then, oa . ob = |oa | |ob | cosT
= |oa | |ob | cos90° = 0 o
Hence, oa . ob = 0 b
o
a
Conversely, of oa . ob = 0, then T = 90°
oo
(b) If a and b are parallel to each other,
i.e., T = 0° or T = 180°
o o |oa | |ob | cos0° = 1
a. b=
and o o = |oa | |ob | cos180° = ab.(-1) = - ab
a. b
oo oo
Conversely, if a . b = ± ab, T = 0° or 180°, a and b are parallel to each other.
Some Important Results
a. If |oa | = a and |ob | = b, oo
a . b = abcosT.
oo (i) |oa |= 0 (ii) |ob | = 0 or o o
b. If a . b = 0, a A b
oo
c. If T = 0°, the value of a . b = ab is maximum.
oo
d. If T = 180°, the value of a . b = - ab is minimum.
oo oo
e. a . b = b . a .
oo o o oo
f. a . (k b ) = (k a ) . b = k( a . b ), where k z 0. k R.
g. o o = |oa | |oa | cos0 = a.a. =a2
a. a
Also,
(i) o o = |oi | |oi | cos0° = 1.1.1 = 1
i. i
(ii) o o = |oj | |oj | cos0° = 1.1.1 = 1
j. j
(iii) o o = |oi | |oj | cos90° = 1.1.0 = 0
i. j
o . o = |oj | |oi | cos90° = 1.1.0 = 0
(iv) j i
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Angle between Two Vectors
Let us consider two points A(a1, a2) and B (b1, b2) in the plane. Then,
position vector of A = OA = oa = a1
a2
= ob = b1
Position vector of B = OB b2
Magnitudes of oa and ob are
| OA | = OA = a = |oa |
| OB | = OB = b = |ob |
Let XOA = E, XOB = D and AOB = T. Then, AOB = D – E = T.
Draw perpendiculars AM and BN from A and B to the x-axis. Then,
OM = a1, MA = a2, ON = b1 and NB = b2.
From the right-angled triangle OMA,
cosE = OM = aa1 ? a1 = a cos E
OA ? a2 = a sinE
aa2
sinE = MA =
OA
Similarly from the right-angled triangle ONB,
b1 = b cos D and b2 = b sin D
Now, a1b1 + a2b2 = a cos E b cosD + a sin E b sin D
= ab cos(D – E) = |oa | |ob | cos T ................(i)
But, by the definition of scalar product of two vectors, oa and ob
oa · ob = | oa | |ob | cos T .............(ii)
Now, from (i) and (ii) oa · ob = a1b1 + a2b2.
This result leads us to define scalar product of two vectors in another way.
Let oa = a1 and oa = b1 be two vectors. Then the scalar product of oa and ob is denoted
a2 b2
by oa · ob and is defined by oa · ob = a1 b1
a2 · b2 = a1b1 + a2b2.
Again from (i), |oa | |ob | cos T = a1b1 +a2b2
or, cos T = a1b1 + a2b2 or oa . ob
|oa | |ob | |oa | |ob |
oa ob .
This result gives us angle between two vectors and
cosT = a1b1 + a2b2 or oa . ob
|oa | |ob | |oa | |ob |
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Worked out Examples
oo oo
Example 1. (a) If a = (4,5) and b = (2, 3), find the scalar product of a and b .
Solution:
(b) If |oa |= 4 and o = 3, T = 45°. find o o .
Example 2. b a. b
Solution:
oo
Example 3. (a) Here, a = (4, 5), b = (2, 3)
Solution:
Since (a1, a2). (b1,b2) = a1b1 + a2b2
oo
a . b = (4, 5).(2,3) = 4.2 + 5.3 = 8+15= 23
oo
(b) Here, | a | = 4, | b | = 3, T = 45°
oo o o 1 = 6 2
a . b = | a | | b | cosT = 4.3 cos45° = 12. 2
o oo ooo
If a = 2 i + j and b = i - 2 j , then, find the following:
oo oo
(i) a . b (ii) angle between a and b
o o oo o o
Here, a = 2 i + j , b = i - 2 j
oo o o o o
(i) a . b = (2 i + j ). ( i - 2 j ) = 2.1 + 1.(-2) = 2 - 2 = 0
oo oo
Since a . b = 0, a and b are perpendicular to each other.
o
(ii) Here,| a | = 22 + 12 = 4 + 1 = 5
o
| b | = 12+(-2)2 = 1 + 4 = 5
oo
Let T be the angle between a and b , we get,
cosT = oo = 0 =0
a. b 55
| oa || ob |
? T = 90°
ooo o oo o
(a) If a = 3 i + 4 j and b = -8 i + 6 j , show that a is perpendicular
oo o oo
to b , (b) If a = (2, 4) and b = (4, 8), show that a and b are parallel to
each other.
o o oo oo
(a) Here, a = 3 i + 4 j , b = - 8 i + 6 j
oo oo oo
a . b = (3 i + 4 j ) . (-8 i + 6 j )
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= 3.(-8) + 4.6 = - 24 + 24 = 0
oo o o
Since a . b = 0, a and b are perpendicular to each other.
( ) ( )(b)
o 2 o 4
Here, a = 4 , b= 8
( ) ( )o o 2 . 4 = 2.4 + 4.8 = 8 + 32 = 40
4 8
a. b =
o
| a | = 22 + 42 = 4 + 16 = 20 = 2 5
o
| b | = 42 +82 = 16+64 = 80 = 4 5
oo
Let T be the angle between a and b
Now, cosT oo =2 40 5 = 1 = cos0°
= a. b 5 ×4
oo
| a || b |
? T = 0°
oo
Hence, a and b are parallel to each other.
oo oo
Example 4. (a) Find the value of m if two vectors m i + 7 j and 4 i + 8 j are
Solution:
perpendicular to each other.
( ) ( )o
(b) If OA =
2 o -3 and AOB = 90°, find the value of m.
m and OB = 4
(a) Here, let o = o + o o oo
a mi 7 j, b =4 i+8 j
oo oo
Since a and b are at right angle, a . b = 0
oo oo oo
i.e. a . b = (m i + 7 j ) . (4 i + 8 j )
or, 0 = 4m + 56
or, 4m = - 56 ? m = -14
( ) ( )(b) o o
oo 2 , OB = b= -3 , AOB = T = 90°
Let OA = a = m 4
Then, by definition of dot product,
( ) ( )o o 2 . -3 = 2(-3) + m.4 = - 6 + 4m
m 4
a. b= oo
Since, T = 90° or, a . b = 0
or, - 6 + 4m = 0
or, 4m = 6 ? m= 3
2
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Example 5. If o = oo and o o o oo
Solution: OA 2 i +6 j OB = 2 i+ j find the angle between OA and OB.
Example 6. o o oo o o
Solution: Here, OA = 2 i + 6 j , OB = 2 i + j .
Example 7. oo
Solution: Let T be the angle between OA and OB . Then,
oo oo
a. b OA . OB
cosT = =
| oa || o OoA || o
b | | OB |
oo o o oo
But, OA . OB = (2 i + 6 j ) . (2 i + j )
= 2 × 2 + 6 × 1 = 10
o
| OA | = 22 + 62 = 4 + 36 = 40 = 2 10
o
| OB | = 22 +12 = 4+1 = 5
? cosT = 2 10 5 = 10 2 = 1 = cos45°
10. 10 2
? T = 45°
Hence, the angle between them is 45°.
If oo = 12, T = 45°, | oa | = 4, find | ob |.
a. b
Here, o o = 12, o = 4
a. b | a|
By definition of dot product of two vectors,
o . o = | oa | | ob | cosT
a b
o
or, 12 = 4 | b | cos45°
or, o 1 o
3=| b| 2 ? b =3 2
oo o o oo
If | a + b | = | a - b | , then, prove that a and b are perpendicular to
each other.
o o oo
Here, | a + b | = | a - b |
Squaring on both side, we get
oo oo
| a + b |2 = | a - b |2
or, (oa + ob )2 = (oa - ob )2 (?|oa |2 = a2)
oo oo
or, a2 + 2 a . b + b2 = a2 - 2 a . b + b2
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oo oo
or, 4 a . b = 0 ? a.b =0
oo
Hence, a and b are perpendicular to each other.
Example 8. If P(0,2), Q(2,5) and R(8,1) are the vertices of a triangle ∆PQR, prove that
Solution: ∆PQR is a right angled triangle
oo o
Here, we have OP = (0,2), QR = (2, 5), OR = (8, 1)
o oo
Now, PQ = OQ - OP = (2, 5) - (0,2) = (2, 3)
o oo
QR = OR - OQ = (8, 1) - (2, 5) = (6, -4)
o oo
RP = OP - OR = (0, 2) - (8,1) = (-8, 1)
oo
PQ . QR = (2, 3) . (6,- 4) = 2.6 + 3.(-4) = 12 - 12 = 0
oo
? PQ . QR = 0, angle between PQ and QR is 90°.
Hence, ∆ PQR is a right angled triangle.
o oo
Example 9. Find the angle made by a = 6 i - 8 j with X - axis.
Solution: oo
We know that unit vectors along X-axis and Y-axis are i and j respectively.
oo
Let b = unit vector along X-axis = (1, 0) = i
o oo
Given a = 6 i - 8 j
oo
Let T be the angle between a and b .
oo o o .oi
a. b (6 i - 8 j )
cosT oo = = 6 = 3
62 + (-8)2 12 + 02 100 50
| a || b |
( )?
T = cos-1 3 = 86.57°
50
o o o
Example 10. If p+ q+ r = (0, 0), | o o o 127 , find the angle
oo p| = 6, | q | =7,| r|=
between p and q .
oo
Solution: Let T be the angle between p and q .
oo o
| p | = 6, | q | = 7, | r | = 127
o oo
Here, p + q + r = 0
oo o
or, p + q = - r
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Squaring on both sides,
oo
( p + q )2 = r2
oo
or, p2 + 2 p . q + q2 = r2
oo
or, 62 + 2| p | | q | cosT + 72 = 127
oo
or, 2 × 6 × 7 × cosT = 127 - 85, where T is angle between a and b .
or, cosT = 42
84
1
or, cosT = 2 = cos60° ? T = 60°
Hence, the angle between two vectors is 60°.
Example 11. Given that o + o and o - o are orthogonal vectors and o o are
(a 2 b) 5a 4b oo a, b
unit vectors. Find the angle between a and b .
oo oo
Solution: Here, a and b are unit vectors. | a | = 1, | b | = 1 .
Given vectors are orthogonal to each other. Hence, their dot product is zero.
o ooo
? ( a + 2 b ).(5 a - 4 b ) = 0
o o o oo o
or, a . (5 a - 4 b ) + 2 b .(5 a - 4 b ) = 0
oo oo
or, 5a2 - 4 a . b + 10 a . b - 8b2 = 0
o o oo
or, 5.1 - 4 a . b + 10 a . b – 8.1 = 0
oo
or, 6 a . b = 3
or, | o | o . cosT = 1 or, 1.1. cosT = 1
a| b| 2 2
or, cosT = 1 or, cosT = cos60°
2
? T = 60°
Hence, angle between them is 60°.
Exercise 10.1
Very Short Questions
1. (a) Define scalar product of two vectors.
(b) If T is the angle between oa and ob , what is the formula of dot product of o and
o a
b?
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(c) If o o = 0, what is the relation between o and ob ?
a. b a
(d) If o = o y1 o and o o y2 o find o o
a x1 i + j b= x2 i + j, a. b.
( ) ( )(e) If oa =a1 and o = b1 , find the value of oa . ob .
a2 b b2
oo
(f) Under which condition two vectors a and b are perpendicular to each other ?
oo
2. Find the dot product of a and b in following cases:
oo
(a) | a | = 2, | b | = 3, angle between them (T) = 60°
oo
(b) | a | = 2, | b | = 5, angle between them = 90°
oo
(c) | a | = 15, | b | = 25, angle between them = 45°
oo
3. Find the dot product of the vectors a and b in the following cases:
( ) ( )(a) a = 4 , b = 1
o 3o 2 oo oo o o
(b) a = i + 2 j , b = 3 i - j
o o oo o o o oo o
(c) a = 4 i + 2 j , b = - i + 2 j (d) a = 10 i , b = 5 j
oo o o oo oo oo
4. If a = (2, 1), b = (2, –3), find (i) a . b (ii) b . a (iii) show that a . b = b . a
o 2o 4o 2
( ) ( ) ( )5. If a = 3 , b = 2 , c = -3 , find:
(a) a2 (b) b2 (c) c2
oo oo oo
(d) a . b (e) b . c (f) a . c
Short Questions
6. Find the angles between given pair of vectors:
o o oo o o oo
(a) | a | = 4, | b | = 5, a . b = 10 (b) | a | = 3, | b | = 5, a . b = 7.5
( ) ( )(c) a = -3 , b = -2
o 2o 1 oo
(d) p = (3, 4), q = (-4, 3)
o o oo o o
(e) p = 4 i + 2 j , q = - i + 2 j
7. Show that following pair of vectors are parallel:
o 2o 4 ( ) ( )o 1 o -3
(b) p -3 , q = 9
( ) ( )(a) p = 3 , q = 6
o o oo o o
o o oo o o (d) a = i + 5 j , b = -2 i - 10 j
(c) a = 2 i - 3 j , b = 4 i - 6 j
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8. Show that pair of vectors are perpendicular to each other in the following cases :
oo oo
(a) a = (3, 6), b = (6, -3) (b) a = (1, 2), b = (-2, 1)
( ) ( )(c) p = 2 i - 3 j , q = 2 j + 3 i (d) p = 4 , q = 3 -4
o o oo o o o 3o
9. Find the value of k if each pair of the following vectors are orthogonal to each other:
oo
(a) a = (2, -3), b = (k, 4)
o o oo oo
(b) a = (4 i + k j ) , b = (3 i - 6 j )
o o oo o o
(c) a = 3 i - 2k j , b = 10 i - 6 j
o o oo o o
(d) p = 2k i +4 j , b = 3 i +2 j
Long Questions
oo oo o
10. Express PQ and RS in the form of x i + y j and show that PQ is perpendicular to
o
RS in the following cases.
(a) P(3, 5), Q(4, 6), R(7, 6), S(8,5) (b) P(3, -2), Q(5, 1), R(-1, 4), S(2, 2)
11. (a) In ∆PQR if o = o - o , o = o o prove that ∆PQR is a right
PQ 5i 9j QR 4i + 14 j
angled triangle.
o o o o oo
(b) If PQ = – 3 i + 4 j and PR = – 7 i + j prove that ∆PQR is an isosceles right
angled triangle.
oo
12. (a) Find the angle between 4 i – 3 j with x-axis.
oo
(b) Find the angles between 2 i + j and y - axis.
oo ooo
13. (a) Show that the angle between two vectors a and c is 90° if a + b + c = (0, 0),
ooo
| a | = 3, | b | = 5, | c | = 4.
o o ooo o o
(b) Find the angle between a and b if a + b + c = 0, | a | = 6, | b | = 7,
o
| c|= 127 o o o
ooo
(c) If a + b + c = 0, | a | = 3, | b | = 5, | c | = 7, show that the angle between
oo
| a | and | b | is 60°.
oo oo oo
14. If ( a + b ) and (2 a - b ) are orthogonal vectors, a and b are unit vectors, find the
oo
angle between a and b .
oo o o oo
15. If | a - 3 b | = | a + 3 b | , prove that a and b are orthogonal vectors.
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2. (a) 3 (b) 0 (c) 375 2
3. (a) 10 (b) 1 2
4. (i) 1 (ii) 1
(d) 14 (c) 0 (d) 0
(c) 13 (b) 60°
6. (a) 60° 5.(a) 13 (b) 20
(b) 2
9. (a) 6 (b) 63.43° (e) 2 (f) -5
12. (a) 71.56°
(c) 7.13° (d) 90° (e) 90°
(c) - 5 (d) - 4
2 3
13.(b) 60° 14. 180°
10.2 Vector Geometry
Review : A
From the given figures, answer the following questions :
oo
(a) Find the sum of AB and BC in the triangle ABC.
oo o B (i) C
(b) In ∆ABC, can we write BC + CA + AB = 0 ? S
R
oo
(c) In the given rectangle PQRS, write a single vector representing PQ + QR .
oo P
(d) Is PQ . QR = 0?
oo Q 10cm
(e) What is difference between PQ . QR and PQ.QR ? (ii)
(f) State the triangle law and parallelogram laws in vector addition.
Triangle Law of Vector Addition
"If the magnitude and direction of two vectors are represented by two sides of a triangle
taken in order, then the magnitude and direction of their sum is given by the third side taken
in reverse order."
Parallelogram Law of Vector Addition
"If two adjacent sides of a parallelogram through a point represent two vectors in magnitude
and direction, their sum is given by the diagonal of the parallelogram through the same
point in magnitude and direction."
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10.3 Theorems on Vector Geometry
THEOREM 1
Mid Point Theorem
oo
If a and b are the position vectors of two points A and B and M is the mid-point of the
o o o
line segment AB, the position vector of M is OM = 1 (a + b)
2
Proof : Let O be the origin and AB be a line segment.
Position Vector of A = o = o oB
OA a b
Position Vector of B = o = o
OB b
M is the mid-point of AB. O M
A
Then by using triangle law of vector addition, we have, o
a
o oo o 1 o
OM = OA + AM = OA + 2 AB
= o 1 ( o - o
OA + 2 OB OA )
= o 1 o - o
a+ 2 (b a)
= oo o oo
2 a+ b- a a +b
2 = 2
? o 1 o o Proved.
OM = 2 (a + b)
Alternative Method
Since, M is the mid point of AB
oo
AM = MB
oo oo
or, OM – OA = OB – OM
ooo
or, 2 OM = OA + OB
? o 1 o + o Proved.
OM = 2 (a b)
Note :
If A(x1, y1) and B(x2, y2) are any two points and M is the mid point of line segment AB, then
the position vector of M is given
( )o
OM =
x1 + x2 , y1 + y2
2 2
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THEOREM 2
Section formula for internal division.
oo
If a and b are the position vectors of two points A and B respectively and P divides the
line segment AB internally in the ratio m:n, the position vector of vector of P is
oo oo
OP = p = mb + na
m+n
Proof : Let O be the origin and AB be a line segment. P divides AB internally in the ratio
m:n. B
oo o n
Here, position vector of A = OA = a bo P
position vector of B = o = o p
OB b
position vector of P = o = o m
OP p
O oA
Here, AP = m a
PB n
o oo
Now, AP = m ( AP and PB are in the same direction)?
o n
PB Proved.
oo
or, n AP = m PB
oo oo
or, n( OP - OA ) = m ( OB - OP )
oo oo
or, n( p - a ) = m b - m p
oo oo
or, m p + n p = m b + n a
oo
o mb + na
? p = m + n
Note :
o 1 ( o o where o = o o = o o = o
If P is the mid point of AB, then p = 2 a + b ), OP p, OA a, OB b
THEOREM 3
Section formula for external division
oo
If a and b are position vectors of two points A and B respectively and the point
P divides the line segment externally in the ratio m:n, the position vector of P is
oo
o mb – na
p = m–n
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Proof : Let O be the origin and AB be a line segment. The point P divides line segment AB
in ratio m:n externally. P
Here, position vector of A = o o o
OA = a po
position vector of B = o o bB
OB = b
position vector of P = o o O
OP = p
o
Now, AP = m a A
BP n
o
or, AP = m oo?
o n ( AP and BP have same direction)
BP
oo
or, m BP = n AP
oo oo
or, m( OP - OB ) = n ( OP - OA )
oo oo
or, m p - m b = n p - n a
o oo
or, (m - n) p = m b - n a
oo
o mb – ma
? p = m–n Proved.
THEOREM 4
Centroid Formula
oo o
If a , b , and c are position vectors of the vertices A, B and C of ∆ABC, the position
o o o o o
vector of centroid of triangle is OG = g = 1 ( a+ b+ c)
3
Proof : Let O be the origin and A,B and C the vertices of ∆ABC. o
a A
Then, oo O o
position vector of A = OA = a g
position vector of B = o = o o Go
OB b b c
position vector of C = o = o DC
OC c
Consider a median AD and G the centroid of the triangle, B
G divides AD into 2:1 ratio. i.e. AG : GD = 2:1 = m:n
Now, o 1 ( o + o = 1 ( oo
OD = 2 OB OC ) 2 b +c)
o oo 2× 1 ( oo o 1 o o o
OG m OD + n OA = 2 b +c) + 2. a 3 a b + c)
2+1
? = m+n = ( + Proved.
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Worked out Examples
oo oo
Example 1. The position vectors of A and B are 6 i + 2 j and 2 i + 4 j respectively.
Solution:
If M is the mid-point of line segment AB, find the position vector of M.
Example 2.
Solution: Let O be the origin.
oo o o
Then, position vector of A = OA = a = 6 i + 2 j
position vector of O ob B
oo o o M
oa
B = OB = b = 2 i + 4 j
M is the mid-point if line segment AB.
? position vector of M o 1 ( o + o ) A
= OM = 2 a b
= 1 ( 6 o + 2 o + 2 o o
2 i j i +4 j )
= 1 ( o + o ) = 4 o o
2 8i 6j i +3 j
oo oo
The position vectors of A and B are respectively 4 i +7 j and 6 i - 2 j .
Find the position vector of C and D if
(a) C divides AB in 2:1 ratio internally. B
(b) D divides AB in 3:2 ratio externally. 1
C
(a) Let O be the origin . 2
Then, ooo A
positive vector of A = OA = 4 i + 7 j
positive vector of B = o = o o . O
OB 6i –2 j
C divides AB in ratio 4:2 internally. i.e. m:n = 4.2
oo
m OB + n OA
Now, position vector of C = m+n
2(6 o – 2 o )+ 1(4 o + 7 o )
i j i j
= 2+1
)o o o o
= 12 i - 4 j + 4 i +7 j
3
oo
16 i +3 j
= 3
= 16 o o
3 i +j
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(b) Here, D divides AB in 3:2 ratio externally. i.e. m:n = 3:2
oo
m OB – n OA
Now, position vector of D = m+n
3(6 o – 2 o )- 2(4 o + 7 o )
i j i j
= 3-2
)o o o o
= 18 i – 6 j - 8 i - 14 j
1
oo
= 10 i – 20 j
Example 3. If A(4, 4), B(3, 2), and C(3, 4) are the vertices of ∆ABC, find the position
Solution: vector of centroid the triangle.
Let O be the origin. Then we have,
o
position vector of A = OA = (4, 4)
o
position vector of B = OB = (3, 2)
o
position vector of C = OC = (3, 4)
Now, position vector of centroid G of ∆ABC is given by
o = 1 o oo
OG 3 { OA + OB + OC }
= 1 {(4, 4) + (3, 2) + (3, 4)}
3
1
= 3 (10, 10)
= 10 , 10
3 3
Exercise 10.2
Short Questions
oo
1. Let the position vectors of A and B be respectively a and b , AB is the line segment
joining them. Then, write the formula in the following conditions:
(a) position vectors of the mid point M of AB.
(b) position vector of the point P on AB which which divides AB in ratio of m:n
internally.
(c) position vector of the point P of AB which divides AB in ratio of m:n externally.
2. (a) The position vectors of A and B are respectively 4 and 6 , find the
( ) ( )5 3
position vector of the mid point of the line segment AB.
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oo oo
(b) The position vectors of P and Q are respectively. (4 i + 2 j ) and (3 i + 3 j ) ,
then find the position vector of the mid point of line segment PQ.
oo oo
(c) The position vector of M and N are respectively (3 i + 5 j ) and (- i + 3 j ), if
oo
MP = PN , find the position vector of P.
oo oo
(d) The position vectors of A and B are (2 a - 3 b ) and (5 a - 4 b ) respectively. Find
the position vector of mid point of AB.
oo
3. (a) If the position vector of the mid point of the line segment AB is (3 i - 2 j ), where
oo
the position vector of B is (5 i + 2 j ), find the position vector of A.
oo
(b) The position vector of the mid point of the line segment PQ is (4 i + 2 j ), where
oo
the position vector of P is (3 i - 7 j ). Find the position vector of Q.
o o o oo o
4. (a) The position vectors of A and B are respectively a =3 i +4 j and b = i - 2 j . If
C divides AB in the ratio of 3:2 internally, find the position vector of C.
oo oo
(b) The position vectors of A and B are respectively (4 i + 6 j ) and (2 i + 3 j ) .
Find the position vector of C which divides AB in the ratio 2:3 internally.
(c) The position vectors of A and B are 2 o - 3 o oo
a b and 5 a – 4 b respectively. Find
the position vector of the point which divides AB in ratio 3:2 internally.
oo oo
5. (a) The position vectors of A and B are respectively (3 i - 2 j ) and (3 i + 6 j ).
Find the position vector of P which divides AB in ratio 2:3 externally.
oo oo
(b) The position vector of A and B are respectively ( i + j ) and (3 i + 5 j ). Find
the position vector of P which divides AB in ratio of 2:1 externally.
6. (a) If the position vectors of the vertices A, B, and C of ∆ABC are respectively
o o oo oo
(3 i + 5 j ), (5 i - j ), and ( i + 8 j ), find the position vector of centroid
of the triangle.
oo oo
(b) The position vectors of P, Q and R of ∆PQR are respectively (3 i +4 j ), (4 i +5 j ),
oo
and (5 i + 6 j ), find the position vector of centroid of the triangle.
(c) In ∆LMN, o o o o = o + o and the position vector of
OL = 4i -5 j , OM 6i 4j
o oo o
centroid G, OG = 2 i + j . Find ON.
o o oo oo
(d) In ∆ABC, OA = 3 i - 5 j , OB = - 7 i + 4 j and centroid G,
o oo o
OG = 2 i + 6 j , find OC .
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(e) Find the position vector of centroid G of ∆ABC, with vertices A(-1, -1), B(-1, 5),
and C(5, 2).
Long Questions
oo
7. (a) If a and b are the position vectors of points A and B respectively. Find the
oo
position vector of C in AB produced such that AC = 3 BC
(b) OABC is a parallelogram. P and Q divide OC and BC in the ratio of
o oo o o
CP:PO = CQ:QB=1:3. If OA = a , OC = c , find the vector PQ and show that
PQ//OB.
8. (a) In the given figure if A
o 3 o oo
CB 2 AB AC = 3 AD
o = , then show that : 2 o +
o PQ ,
CD 1 o PB D C
(b) In the given figure PA = 6 p A
then show that o
Oa
( )o o
qQ
a=
1 oo
6 5 p+ q
9. (a) In ∆ABC, the medians AD, BE and CF are drawn from the vertices A, B and C
ooo
respectively. Then prove that , AD + BE + CF = (0, 0). A
(b) If G is centroid of ∆ABC, FE
then prove that G
o oo
GA + GB + GC = O B DC
Project Work
10. State the triangle law of vector addition. Is this theorem necessary to prove mid-point
theorem and section formulas in vector geometry? Illustrate with examples.
11. State differences between theorems proved in vector geometry and plane geometry.
Illustrate with examples.
2. (a) 5 (b) 7 i + 5 j (c) i + 4 j (d) 7 a – 7 b
4 2 2 2 2
9 2
3. (a) i – 6 j (b) 5 i + 11 j 4.(a) 5 i + 5 j
(b) 16 i + 24 j (c) 19 a – 18 b 5.(a) 3 i – 18 j (b) 5 i + 9 j
5 5 5 5 (d) 10 i + 19 j
6. (a) 3 i + 4 j (b) 4 i + 5 j (c) -4 i + 4 j
(e) (1, 2) 7.(a) 3 b – a
2
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10.4 Theorems Related to Triangles
THEOREM 5
The line segment joining the mid-point of two sides of a triangle is parallel to the third side
and is half of it. (prove vectorically).
Solution:
Let ABC be a triangle in which D and E are the mid-points of AB and AC respectively, EF is
the line segment joining E and F. A
To prove : o = 1 o o // o
DE 2 BC , DE BC
Proof : DE
o oo B C
Here, DE = DA + AE (by triangle law of vector additional)
or, o 1 o 1 o (E and F are mid-points of AB and AC)
DE = 2 BA + 2 AC
or, o 1 oo
DE = 2 ( BA + AC )
o 1 o
DE = 2 BC
oo o o 1
? DE // BC ( DE =k BC , k = 2 )
o o
? DE = 1 BC Proved.
2
THEOREM 6
The median to the base of an isosceles triangle is perpendicular to the base.
Solution:
Let ∆ABC be an isosceles triangle with AB = AC, D be the mid point of base BC. AD is
median drawn from vertex A to BC.
To prove : AD A BC A
Proof :
oo o
1. BC = BA + AC (by triangle law of vector addition)
o oo
AD = AB + AC (by mid-point theorem)
( )2. 1 BD C
2
oo
AC + AB
( )=1
2
oo o o oo
BC . AD = AC AB ) . AB + AC
( ( )3. - 1
2
oo oo oo
AC - AB AC + AB AC 2 - AB 2
( ) ( ) [( ) ( ) ]=1 . = 1
2 2
1 1
= 2 [AB2 - AB2] = 2 .0 =0 ( ? AB = AC)
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oo
Since, BC . AD = 0.
oo
? AB is perpendicular to BC . Proved.
THEOREM 7
The mid-point of the hypotenuse of a right angled triangle is equidistant from its vertices
Solution: A
Let ∆ABC be a right angled triangle with ABC = 90°,
D the mid point of hypotenuse AC. i.e. AD = DC D
To prove : AD = BD = CD
Proof :
oo o B C
1. In ∆ABD, AB = AD + DB (by triangle law of vector addition)
o o o oo o o o o
2. In ∆BDC, BC = BD + DC = BD + AD = AD + BD ( AD = DC )
3. Since B = 90°, so we can write,
oo
AB . BC = 0
oo o o
or, ( AD + DB ) . ( AD + BD ) = 0
or, ( o - o ) . ( o + o ) = 0
AD BD AD BD
or, ( o )2 = ( o )2
AD BD
or, AD2 = BD2
or, AD = BD
4. Again, D is mid-point of AC i.e. AD = DC
5. From (3) and (4), we get
AD = BD = CD Proved.
Worked out Examples
Example 1. Prove by vector method that the perpendicular drawn from the vertex to the
Solution: base of an isosceles triangle bisects the base.
Let ∆ABC be an isosceles triangle with AB = AC. AD is drawn perpendicular
to BC. A
To prove : BD = DC
Proof :
1. Since AD A BC B DC
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oo oo
then, AD . BD = 0 or, AD . DC = 0
ooo
2. AB = AD + DB (by using triangle law of vector addition)
ooo
AC = AD + DC
oo ( ABC is an isoceles triangle)
3. | AB | = | AC |
oo oo
or, | AD + DB | = | AD + DC |
squaring on both sides, we get
( o o )2 = ( o + o )2
AD + DB AD DC
oo oo
or, AD2 + 2 AD . DB + DB2 = AD2 + 2 AD . DC + DC2
or, 2.0 + DB2 = 2.0 + DC2 (by using 1)
or, DB2 = DC2 BD = DC Proved.
Example 2. In the given ∆PQR, RM and QN are medians of ∆PQR such that RM = QN,
Solution: then prove vectorically that ∆PQR is an isosceles triangle.
In ∆PQR, medians RM = QN
To prove : ∆PQR is an isosceles P
Proof :
We have, QN = RM M
i.e
| o | = | o | N
QN RM R
squaring on both sides, we get, Q
oo
| QN|2 = | RM|2
oo oo
or, ( QP + PN )2 = ( RP + PM )2
oo oo
or, QP2 + 2 QP . PN + PN2 = RP2 + 2 RP . PM +PM2
QP2 +( ) ( )or,142=QPo43QP 2.P-R122PoPoQR.+ 1 PR 2 = RP2 +2 o 1 o + 1 PQ 2
o 2 PR2 - 1 PR2 o RP . 2 PQ 2
or, PR = 4
PQ2 -or, - PR o
. PQ
3 PQ2
4
or, PQ2 = PR2
oo
or, | PQ |2 = | PQ |2
? PQ = PR.
Hence, ∆PQR is an isosceles. Proved.
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Exercise 10.3 S T
Q
Short Questions R
1. (a) In the given figure S and T are the mid points of sides P
PQ and PR. Write the relation between ST and QR. Q MR
(b) In the given ∆PQR, PQ = PR, PM is a median. Write the P
relation between PM and QR. M
(c) In the given ∆PQR, M is the mid-point of PR and
QR
PQR = 90°. Write the relations among PM, MR and
QM.
(d) In the given ∆PQR if PM = MR = QM. Find the relation
between PQ and QR.
oo
(e) If O is origin and OA = a ,
o oo o oo A
OB = b , OC = c . Write the relation of a , b ,
oa
oo og
c and g , where G is the centroid of the triangle. O
G
ob C
oc
B
D
Long Questions P
2. (a) In ∆PQR, PQR = 90°.
Prove that PR2 = PQ2 + QR2
QR
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(b) ∆ABC is a right angled triangle, B = 90°. A
Let C = T, AC = h, BC = b, AB = p, then prove B
that : h2 = p2 + b2
CT
A
(c) In the figure ∆ABC is an isosceles triangle. AD is a
oo
median, show that AD . BC = 0
B DC
A
3. In triangle ABC, M and N are the mid points of AB
and AC respectively. Then, prove vertically that MN
(a) MN//BC
(b) o = 1 o PB C
MN 2 BC
4. In the given triangle, PM is a median of an isosceles,
triangle PQR with PQ = PR. Then show that PM is
perpendicular to QR, vertically. Q MR
5. Prove vectorically that the mid point of hypotenuse of a right angled triangle is
equidistant from its vertices.
6. Prove vectorically that the line joining the vertex and the mid point of P
the base of an isoscles triangle is perpendicular to the base.
7. ∆PQR is an isosceles triangle with PQ = PR and PM A QR. Then
vectorically show that QM = RM. Q MR
oo o P
8. If a , b , and c are the position vectors of the vertices
G
A, B, C of ∆ABC respectively. Prove vectorically that the position MR
ooo
a+b+ c
o 1 Q
3
( )vector of centroid G of ∆ABC is g =
1. (a) ST // QR, o = 1 o (b) PM A QR (c) PQ A QR, PQR = 90°
ST 2 QR
(d) PM = RM = QM (e) o = 1 o + o + o
OG 3 (a b c)
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10.5 Theorem on Quadrilateral and Semi - circle
THEOREM 8
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is
a parallelogram.
Solution: AS D
In the figure, P,Q,R and S are the mid-points of sides AB, BC, CD,
and AD respectively of a quadrilateral ABCD. Joining P,Q,R, and S P R
successively a quadrilateral PQRS is formed.
To prove : PQRS is a parallelogram. B QC
Construction : Join BD.
Proof :
1. In ∆ABD, o = 1 o
PS 2 BD (In ∆ABD, P and S are the mid-points of AB and AD)
2. In ∆BCD, o = 1 o (In ∆BCD, Q and R are the mid-points of BC and CD)
QR 2 BD
oo
3. PS = QR = [From statement (1) and (2)]
4. | PoS| = | QoR|, PQ // SR (Magnitudes and directions of equal vectors o and o equal)
PS QR
o o oo
5. Similarly, | SR | = | PQ | , SR // PQ . (as in statement (4))
6. ? PQRS is a parallelogram. (from (4) and (5) opposite sides are parallel).
THEOREM 9
The diagonals of a parallelogram bisect each other.
Solution: O
M
In the figure ABCD is a parallelogram. Diagonals D C
AC and BD are drawn. B
Let the mid point of diagonal BD be M.
To prove : M is the mid-point of AC. A
Construction : Let the position vectors of A,B,C,D
oooo o
and M be OA , OB , OC , OD and OM
respectively.
Proof : o o o
OM = ( OD + OB )
1. In ∆ OBD, 1 (by mid-point theorem)
2
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or, o = 1 o + o + o (In ∆ OAD, o = o o
OM 2 ( OA AD OB ) OD OA + AD )
or, o 1 o o o
OM = 2 ( OA + OB + BC )
? o = 1 o o ooo
OM 2 ( OA + OC ) (In ∆ OBC, OB + BC = OC )
2. ? M is the mid-point of AC. (By mid-point theorem o = 1 o o
OM 2 ( OA + OC ))
3. Hence, the diagonals the parallelogram bisect each other C from statement (1) and
(2). Proved.
Alternative Method
Let O be the origin. O
o ooo D C
Then OA , OB , OC , OD are the position vectors of A B
o oo o
A, B C and D respectively. Let OA = a , OB = b ,
oo oo
OC = c and OD = d
Proof
1. By using mid-point theorem, position vector of mid point of diagonal
o o o o
BD = 1 ( b + d ). Position vector of mid-point of diagonal AC = 1 ( a + c ).
2 2
oo
2. Now, AB = DC (opposite sides of a parallelogram)
o o oo
or, OB - OA = OC - OD
oo oo
or, b - a = c - d
oo oo
b+ d a+ c
or, 2 = 2 (dividing both sides by 2)
3. Position vector of mid-point of BD = position vector of mid point of AC, [from (2)].
4. Hence, the diagonals of a parallelogram bisect each other. Proved.
THEOREM 10
The diagonals of a rhombus bisect each other at right angle. Prove it vectorically.
Solution : D C
B
In the figure, let ABCD be a rhombus with diagonal AC and BD.
Then AB = BC = AD = DC.
To Prove : AC A BD, mid-point of AC = mid-point of BD. A
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vedanta Excel In Opt. Mathematics - Book 10
Proof :
ooo (by using triangle law of vector addition)
1. In ∆ ABC, AC = AB + BC
2. In ∆ABD, (by triangle law of vector addition)
o oo oo
BD = BA + AD
oo ( AD = BC )
= BA + BC
oo
= BC – AB
oo
3. Taking dot product of BD and AC ,
oo oo o o
BD . AC = ( BC - AB ) . ( BC + AB )
oo
= BC2 – AB2 ( | BC | = | AB | )
= BC2 – BC2 = 0
4. BD A AC.
o oo oo oo o
5. Let OA = a , OB = b , OC = C , OD = d .
oo
a+ c
Position vector of the mid-point of AC = 2
oo
b+ d
Position vector of the mid-point of BD = 2
oo
6. AB = DC (Opposite sides of parallelogram)
oo oo
or, OB - OA = OC - OD
oo o o
or, b - a = c - d
oooo
? a+ c=b+d
oo oo
a+ c b+ d
? 2 = 2
i.e., Mid-point of diagonal AC = Mid-point of diagonal BD.
7. Hence, the diagonals of rhombus bisect each other at right angle. (From statement (4)
and (6).) Proved.
THEOREM 11
Prove vectorically that the diagonals of a rectangle are equal. D C
Solution: B
Let ABCD be a rectangle.
AC and BD are diagonals of the rectangle. A
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To prove : | o | = | o |
AC BD
Proof :
1. In ∆ABC, o = o + o (By triangle law of vector addition)
AC AB BC
o oo
or, ( AC )2 = ( AB + BC )2
oo oo o
or, ( AC )2 = ( AB )2 + 2. AB . BC + ( BC )2
oo
Since AB A BC, AB . BC = 0,
? AC2 = AB2 + BC2
2. In ∆ABD, o = o + o
BD BA AD
or, o = (- o ) + o . ( o = o )
BD AB BC AD BC
oo oo o
( BD )2 = (- AB )2 + 2( AB ). BC + ( BC ) 2
ooo
? ( DB )2 = ( AB )2 + ( BC )2
? DB2 = AB2 + BC2
3. From (1) and (2), we get.
AC2 = BD2
or, AC = BD
oo
? | AC | = | BD | Proved.
THEOREM 12
Prove vectorically that the angle at semi - circle is a right angle.
Solution : Let O be the centre of the circle and AB be a diameter.
ACB angle at semi-circle. OC is joined. C
To prove : ACB = 90°
Proof : AO B
1. By triangle law of vector addition,
oo o
AC = AO + OC
o oo
CB = CO + OB
oo oo
= CO + AO ( AO = OB radii of the same circle)
oo
= AO - OC
oo
2. Taking dot product of AC and CB , we get,
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vedanta Excel In Opt. Mathematics - Book 10
oo o o o o
AC . CB = ( AO + OC ). ( AO - OC )
oo
= ( AO )2 - ( OC )2
= AO2 - AO2 (OA = OC = OB, radii of same circle)
=0
3. Since o . o = 0, o A o
AC CB AC CB
Thus, shows that ACB = 90° Proved.
Exercise 10.4
Short Questions
1. (a) In the adjoining figure P, Q, R, S are the mid-points of D SC
sides AD, AB, BC, and CD respectively. The points P, Q, R R
and S are joined. What type of quadrilateral is PQRS?
(b) What is the measure of circumference angle at semi- P
circle?
(c) Write relations of diagonals in each of the following A Q B
quadrilateral.
(i) Square (ii) Rhombus (iii) Parallelogram
Long Questions
2. In a quadrilateral ABCD, the mid-points of sides AB, BC, CD, and DA are respectively
P, Q, R and S. Prove vectorically that PQRS is a parallelogram.
3. Prove vectorically that the diagonals of parallelogrm PQRS bisect each other.
4. Prove vectorically that the diagonals of rhombus EFGH bisect each other at right angle.
5. Prove vectorically that the diagonals of rectangle PQRS are equal.
6. Prove vetcotically that the angle at semi-circle is right angle. R
7. In the figure O is the centre of the semi-circle. Prove that
PRQ is right angle.
8. If the diagonals of a quadrilateral bisect each other, prove by P OQ
vector method that is a parallelogram.
BLC
9. In the given figure, ABCD is a parallelogram. L and M are the M
mid-points of sides BC and CD respectively. D
Prove that o + o 3 o A R
AL AM = 2 AC S N
Q
10. In the given figure PQRS is a parallelogram M and N are the mid M
points of PS and QR, show that MQNS is a parallelogram. 297
P
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11. If a line is drawn from the centre of a circle to the mid-point of a chord, prove by vector
method the line is perpendicular to the chord.
12. In the given figure PQRS is a trapezium where PS//QR. M and N P S
are the mid-points of PQ and SR respectively. Prove vectorically N
that M
(i) o = 1 o + o Q R
MN 2 ( PS QR ) D C
oo
(ii) MN // QR
13. In the adjoining figure, PQRS is a parallelogram. G is the point O G
of intersection of the diagonals. If O is any point prove that
oooo A B
OA + OB + OC + OD SM R
o = 1
4 NQ
( )OG C
14. In the adjoining figure PQRS is a parallelogram. M and N are
two points on the diagonal SQ. If SM = NQ, then, prove by
vector method by PMRN is a parallelogram. P
oo D
15. ABCD is a parallelogram and O is the origin. If OA = a ,
o oo o o oo o
OB = b , OC = c find OD in terms of a , b and c .
Project Work A B
O
15. Write three statements of vector geometry. Show differences
between them in vector geometry and plane geometry.
1. (a) parallelogram (b) 90° 15. oa – ob + oc
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11Transformations
11.0 Review
Discuss the answers of the following questions:
(a) Name the four fundamental transformations that you studied in class 9 ?
(b) Can transformations be considered as functions?
(c) What is an invariant point in transformation ?
(d) Can you say any common characteristics between an object and image in reflection,
rotation and translation ?
(e) What is the image of point P(2, 4) when it is rotated through 90° about centre (1, 2) ?
(f) Give any two examples of enlargements used in our daily life.
(g) What type of transformation can be explained by rotating a wheel of bicycle ?
(h) What is the distance between the centre and a point on the circumference of a circle
called ?
(i) When an object is shifted through a certain distance in certain direction, what type of
transformation does it represent?
Review of formulae of transformation :
Let us review the transformation formulae that we studied in class 9.
Reflection : Let p(x,y) be any point and p'(x',y') be an image, of P under a reflection.
(a) Reflection on X - axis
P(x, y) X - axis P'(x, -y)
(b) Reflection on on Y-axis
P(x,y) Y - axis P'(-x, y)
(c) Reflection on line y = x
P(x, y) y =x P'(y, x)
(d) Reflection on line y = -x or x = - y
P(x, y) y= - x P'(-y, -x)
(e) Reflection on line x = h
P(x, y) x=h P'(2h - x, y)
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(f) Reflection on line y = k
P (x, y) y=k P' (x, 2k - y)
Translation :
( )a
Let T = b be a translation vector. Then, image of a point P(x, y) is P'(x', y') under the
translation T.
( )P(x, y) T= a P' (x + a, y + b)
b
Rotation :
Let P(x, y) be any point and P'(x1 , y1) be its image under rotation R, about a centre.
(a) Positive Quarter turn or rotation of + 90° about origin.
P(x, y) p'(-y, x)
Positive quarter turn is equivalent to rotation - 270° about the same centre.
(b) Negative quarter turn or rotation of - 90°
P(x, y) P' (y, -x)
Nagative quarter turn is equivalent to rotation + 270° about the same center.
(c) Half turn or rotation of 180° about origin.
P(x, y) P'(-x, -y) , P(x, y) R[(a, b), 180°] P'(2a – x, 2b – y)
Enlargement
Let E[(0, 0) k] denote the enlargement with centre origin and scale factor k. Let P(x ,y) be a
point. Then, its image is given by
P(x, y) E[(o,o), k] P' (kx, ky)
If the centre of enlargement is (a, b) with scale factor k.
P(x, y) E[(a,b), k] P'(k(x - a) + a, k(y - b) +b)
11.1 Composition of Transformations/Combined Transformation
Introduction Y
P(4, 5)
Let us take a point P(4,5), reflect it on X-axis. Then,
P(4, 5) X-axis P'(4, -5)
Again, the image p' (4, -5) of(4, 5) is reflected on Y X' OX
- axis.
P"(–4, –5) P(4, –5)
P'(4, -5) Y-axis P''(-4, -5) Y'
Here, the image P''(-4, -5) is the image of P(4, 5) due
to combined reflections. Such type of transformation
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Vedanta Opt. Maths Book 10 Final (2078)
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