NCERT Solutions Class 12 Mathematics Part II . 674 Pages (668-1342). Free Flip-Book by Study Innovations

Class XII Chapter 10 – Vector Algebra Maths

Question 19:

If are two collinear vectors, then which of the following are incorrect:

A. , for some scalar λ

B.

C. the respective components of are proportional

D. both the vectors have same direction, but different magnitudes
Answer

If are two collinear vectors, then they are parallel.
Therefore, we have:

(For some scalar λ)

If λ = ±1, then .

Thus, the respective components of are proportional.

However, vectors can have different directions.

Hence, the statement given in D is incorrect.

The correct answer is D.

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Class XII Chapter 10 – Vector Algebra Maths
Exercise 10.3

Question 1:

Find the angle between two vectors and with magnitudes and 2, respectively

having .

Answer

It is given that,

Now, we know that .

Hence, the angle between the given vectors and is .

Question 2: .

Find the angle between the vectors
Answer

The given vectors are

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Class XII Chapter 10 – Vector Algebra Maths

Also, we know that .

Question 3:

Find the projection of the vector on the vector .
Answer

Let and .

Now, projection of vector on is given by,

Hence, the projection of vector on is 0.

Question 4:

Find the projection of the vector on the vector .
Answer

Let and .

Now, projection of vector on is given by,

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Class XII Chapter 10 – Vector Algebra Maths

Question 5:
Show that each of the given three vectors is a unit vector:

Also, show that they are mutually perpendicular to each other.
Answer

Thus, each of the given three vectors is a unit vector.

Hence, the given three vectors are mutually perpendicular to each other.
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Class XII Chapter 10 – Vector Algebra Maths
Question 6: .

Find and , if
Answer

Question 7: .

Evaluate the product
Answer

Page 17 of 46

Class XII Chapter 10 – Vector Algebra Maths

Question 8:

Find the magnitude of two vectors , having the same magnitude and such that

the angle between them is 60° and their scalar product is .
Answer
Let θ be the angle between the vectors

It is given that

We know that .

Question 9: .

Find , if for a unit vector
Answer

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Class XII Chapter 10 – Vector Algebra Maths
are such that is perpendicular to ,
Question 10:

If
then find the value of λ.
Answer

Hence, the required value of λ is 8.

Question 11: is perpendicular to , for any two nonzero vectors

Show that
Answer

Hence, and are perpendicular to each other.

Question 12: , then what can be concluded about the vector ?
If Page 19 of 46

Class XII Chapter 10 – Vector Algebra Maths
Answer .
It is given that

Hence, vector satisfying can be any vector.

Question 14:

If either vector , then . But the converse need not be true. Justify

your answer with an example.

Answer

We now observe that:

Hence, the converse of the given statement need not be true.

Question 15:
If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively,
then find ∠ABC. [∠ABC is the angle between the vectors and ]
Answer
The vertices of ∆ABC are given as A (1, 2, 3), B (–1, 0, 0), and C (0, 1, 2).
Also, it is given that ∠ABC is the angle between the vectors and .

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Class XII Chapter 10 – Vector Algebra Maths

Now, it is known that:

.

Question 16:
Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.
Answer
The given points are A (1, 2, 7), B (2, 6, 3), and C (3, 10, –1).

Hence, the given points A, B, and C are collinear.
Page 21 of 46

Class XII Chapter 10 – Vector Algebra Maths

Question 17: form the vertices of a right

Show that the vectors be position vectors of points A, B, and C
angled triangle.
Answer

Let vectors
respectively.

Now, vectors represent the sides of ∆ABC.

Hence, ∆ABC is a right-angled triangle.

Question 18:
If is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ is unit vector if

(A) λ = 1 (B) λ = –1 (C) (D)

Answer .
Vector is a unit vector if

Page 22 of 46

Class XII Chapter 10 – Vector Algebra Maths

Hence, vector is a unit vector if .
The correct answer is D.

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Class XII Chapter 10 – Vector Algebra Maths
Exercise 10.4
Question 1:
and .
Find , if

Answer

We have,

and

Question 2:

Find a unit vector perpendicular to each of the vector and , where

and .

Answer
We have,

and

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Class XII Chapter 10 – Vector Algebra Maths

Hence, the unit vector perpendicular to each of the vectors and is given by the
relation,

Question 3:

If a unit vector makes an angles with with and an acute angle θ with , then
find θ and hence, the compounds of .
Answer
Let unit vector have (a1, a2, a3) components.

Since is a unit vector, .

Also, it is given that makes angles with with , and an acute angle θ with
Then, we have:

Page 25 of 46

Class XII Chapter 10 – Vector Algebra Maths

Hence, and the components of are .

Question 4:
Show that

Answer

Question 5: .
Find λ and µ if Page 26 of 46

Class XII Chapter 10 – Vector Algebra Maths
Answer

On comparing the corresponding components, we have:

Hence,

Question 6: and . What can you conclude about the vectors ?

Given that
Answer

Then, or , or
(i) Either

(ii) Either or , or

But, and cannot be perpendicular and parallel simultaneously.

Hence, or .

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Class XII Chapter 10 – Vector Algebra Maths
Question 7: given as . Then show

Let the vectors

that
Answer
We have,

On adding (2) and (3), we get:

Now, from (1) and (4), we have:
Hence, the given result is proved.

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Class XII Chapter 10 – Vector Algebra Maths

Question 8:

If either or , then . Is the converse true? Justify your answer with an

example.

Answer

Take any parallel non-zero vectors so that .

It can now be observed that:

Hence, the converse of the given statement need not be true.
Question 9:
Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and
C (1, 5, 5).
Answer
The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5), and
C (1, 5, 5).
The adjacent sides and of ∆ABC are given as:

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Class XII Chapter 10 – Vector Algebra Maths
Area of ∆ABC

Hence, the area of ∆ABC

Question 10:
Find the area of the parallelogram whose adjacent sides are determined by the vector

.

Answer

The area of the parallelogram whose adjacent sides are is .
Adjacent sides are given as:

Hence, the area of the given parallelogram is .
Question 11:

Let the vectors and be such that and , then is a unit vector, if
the angle between and is

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Class XII Chapter 10 – Vector Algebra Maths

(A) (B) (C) (D)
Answer
It is given that

.

We know that , where is a unit vector perpendicular to both and

and θ is the angle between and .

Now, is a unit vector if .

Hence, is a unit vector if the angle between and is .

The correct answer is B.

Question 12:
Area of a rectangle having vertices A, B, C, and D with position vectors

and respectively is

(A) (B) 1
(C) 2 (D)
Answer

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Class XII Chapter 10 – Vector Algebra Maths

The position vectors of vertices A, B, C, and D of rectangle ABCD are given as:

The adjacent sides and of the given rectangle are given as:

Now, it is known that the area of a parallelogram whose adjacent sides are

is .

Hence, the area of the given rectangle is
The correct answer is C.

Page 32 of 46

Class XII Chapter 10 – Vector Algebra Maths
Miscellaneous Solutions

Question 1:
Write down a unit vector in XY-plane, making an angle of 30° with the positive direction
of x-axis.
Answer

If is a unit vector in the XY-plane, then
Here, θ is the angle made by the unit vector with the positive direction of the x-axis.
Therefore, for θ = 30°:

Hence, the required unit vector is

Question 2:
Find the scalar components and magnitude of the vector joining the points

.

Answer

The vector joining the points can be obtained by,

Hence, the scalar components and the magnitude of the vector joining the given points

are respectively and .

Question 3:
A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and
stops. Determine the girl’s displacement from her initial point of departure.
Answer

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Class XII Chapter 10 – Vector Algebra Maths

Let O and B be the initial and final positions of the girl respectively.
Then, the girl’s position can be shown as:

Now, we have:

By the triangle law of vector addition, we have:

Hence, the girl’s displacement from her initial point of departure is
.
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Class XII Chapter 10 – Vector Algebra Maths

Question 4:

If , then is it true that ? Justify your answer.
Answer

Now, by the triangle law of vector addition, we have .

It is clearly known that represent the sides of ∆ABC.

Also, it is known that the sum of the lengths of any two sides of a triangle is greater

than the third side.

Hence, it is not true that .
is a unit vector.
Question 5: .

Find the value of x for which
Answer

is a unit vector if

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Class XII Chapter 10 – Vector Algebra Maths

Hence, the required value of x is .

Question 6:
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

.

Answer
We have,

Let be the resultant of .

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors is

Page 36 of 46

Class XII Chapter 10 – Vector Algebra Maths
Question 7:
, find a unit vector parallel to the
If
.
vector
Answer
We have,

Hence, the unit vector along is

Question 8:
Show that the points A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7) are collinear, and find
the ratio in which B divides AC.
Answer
The given points are A (1, –2, –8), B (5, 0, –2), and C (11, 3, 7).

Thus, the given points A, B, and C are collinear.
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Class XII Chapter 10 – Vector Algebra Maths

Now, let point B divide AC in the ratio . Then, we have:

On equating the corresponding components, we get:

Hence, point B divides AC in the ratio

Question 9:
Find the position vector of a point R which divides the line joining two points P and Q

whose position vectors are externally in the ratio 1: 2. Also, show

that P is the mid point of the line segment RQ.

Answer

It is given that .

It is given that point R divides a line segment joining two points P and Q externally in

the ratio 1: 2. Then, on using the section formula, we get:

Therefore, the position vector of point R is .
Position vector of the mid-point of RQ =

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Class XII Chapter 10 – Vector Algebra Maths

Hence, P is the mid-point of the line segment RQ.

Question 10:

The two adjacent sides of a parallelogram are and .

Find the unit vector parallel to its diagonal. Also, find its area.

Answer

Adjacent sides of a parallelogram are given as: and
Then, the diagonal of a parallelogram is given by
.

Thus, the unit vector parallel to the diagonal is

Area of parallelogram ABCD =

Hence, the area of the parallelogram is square units.

Page 39 of 46

Class XII Chapter 10 – Vector Algebra Maths

Question 11:
Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ

are

Answer
Let a vector be equally inclined to axes OX, OY, and OZ at angle α.
Then, the direction cosines of the vector are cos α, cos α, and cos α.

Hence, the direction cosines of the vector which are equally inclined to the axes

are
.

Question 12:

Let and . Find a vector which is

perpendicular to both and , and .
Answer

Let .
Since is perpendicular to both and , we have:

Also, it is given that:

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Class XII Chapter 10 – Vector Algebra Maths

On solving (i), (ii), and (iii), we get:

Hence, the required vector is .

Question 13:

The scalar product of the vector with a unit vector along the sum of vectors

and is equal to one. Find the value of .

Answer

Therefore, unit vector along is given as:

Scalar product of with this unit vector is 1.

Hence, the value of λ is 1.

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Class XII Chapter 10 – Vector Algebra Maths

Question 14:

If are mutually perpendicular vectors of equal magnitudes, show that the vector

is equally inclined to and .
Answer

Since are mutually perpendicular vectors, we have

It is given that:

Let vector be inclined to at angles respectively.
Then, we have:

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Class XII Chapter 10 – Vector Algebra Maths
Now, as .
,

Hence, the vector is equally inclined to .

Question 15:

Prove that , if and only if are perpendicular,

given .
Answer

Question 16: only when
If θ is the angle between two vectors and , then
(A) (B)

(C) (D)
Answer

Let θ be the angle between two vectors and .

Then, without loss of generality, and are non-zero vectors so

that .

Page 43 of 46

Class XII Chapter 10 – Vector Algebra Maths
It is known that .

Hence, when .

The correct answer is B.

Question 17: is a unit

Let and be two unit vectors andθ is the angle between them. Then
vector if

(A) (B) (C) (D)

Answer
Let and be two unit vectors andθ be the angle between them.

Then, .

Now, is a unit vector if .

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Class XII Chapter 10 – Vector Algebra Maths

Hence, is a unit vector if .
is
The correct answer is D.

Question 18:

The value of
(A) 0 (B) –1 (C) 1 (D) 3
Answer

The correct answer is C. when θ isequal to
Question 19:
If θ is the angle between any two vectors and , then

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Class XII Chapter 10 – Vector Algebra Maths

(A) 0 (B) (C) (D) π
Answer

Let θ be the angle between two vectors and .

Then, without loss of generality, and are non-zero vectors, so

that .

Hence, when θ isequal to .

The correct answer is B.

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Page 46 of 46

Class XII Chapter 11 – Three Dimensional Geometry Maths

Exercise 11.1

Question 1:
If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction
cosines.
Answer
Let direction cosines of the line be l, m, and n.

Therefore, the direction cosines of the line are

Question 2:
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer
Let the direction cosines of the line make an angle α with each of the coordinate axes.

∴ l = cos α, m = cos α, n = cos α

Page 1 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes,

are

Question 3:
If a line has the direction ratios −18, 12, −4, then what are its direction cosines?
Answer
If a line has direction ratios of −18, 12, and −4, then its direction cosines are

Thus, the direction cosines are .

Question 4:
Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.
Answer
The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).
It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2),
are given by, x2 − x1, y2 − y1, and z2 − z1.
The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3.
The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6.
It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are
proportional.
Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B,
and C are collinear.

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Class XII Chapter 11 – Three Dimensional Geometry Maths

Question 5:
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (−
1, 1, 2) and (− 5, − 5, − 2)
Answer
The vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.
Therefore, the direction cosines of AB are

The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4.
Therefore, the direction cosines of BC are

i.e.,

The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., −8, −10, and 2.
Therefore, the direction cosines of AC are

Page 3 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

i.e.,

Page 4 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

Exercise 11.2

Question 1:
Show that the three lines with direction cosines

are mutually perpendicular.
Answer
Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each
other, if l1l2 + m1m2 + n1n2 = 0

(i) For the lines with direction cosines, and , we obtain

Therefore, the lines are perpendicular. and , we obtain
(ii) For the lines with direction cosines,

Therefore, the lines are perpendicular.
(iii) For the lines with direction cosines, and , we obtain

Page 5 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

Therefore, the lines are perpendicular.
Thus, all the lines are mutually perpendicular.

Question 2:
Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line
through the points (0, 3, 2) and (3, 5, 6).
Answer
Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line
joining the points, (0, 3, 2) and (3, 5, 6).
The direction ratios, a1, b1, c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and
−4.
The direction ratios, a2, b2, c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.
AB and CD will be perpendicular to each other, if a1a2 + b1b2+ c1c2 = 0
a1a2 + b1b2+ c1c2 = 2 × 3 + 5 × 2 + (− 4) × 4
= 6 + 10 − 16
=0
Therefore, AB and CD are perpendicular to each other.

Question 3:
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the
points (−1, −2, 1), (1, 2, 5).
Answer
Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through
the points, (−1, −2, 1) and (1, 2, 5).
The directions ratios, a1, b1, c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and
−4.
The direction ratios, a2, b2, c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4,
and 4.

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Class XII Chapter 11 – Three Dimensional Geometry Maths

AB will be parallel to CD, if

Thus, AB is parallel to CD.

Question 4:
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to

the vector .

Answer

It is given that the line passes through the point A (1, 2, 3). Therefore, the position

vector through A is

It is known that the line which passes through point A and parallel to is given by
is a constant.

This is the required equation of the line.

Question 5:
Find the equation of the line in vector and in Cartesian form that passes through the

point with position vector and is in the direction .

Answer

It is given that the line passes through the point with position vector

Page 7 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

It is known that a line through a point with position vector and parallel to is given by
the equation,

This is the required equation of the line in vector form.

Eliminating λ, we obtain the Cartesian form equation as

This is the required equation of the given line in Cartesian form.
Question 6:
Find the Cartesian equation of the line which passes through the point
(−2, 4, −5) and parallel to the line given by
Answer
It is given that the line passes through the point (−2, 4, −5) and is parallel to

The direction ratios of the line, , are 3, 5, and 6.

The required line is parallel to
Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0
It is known that the equation of the line through the point (x1, y1, z1) and with direction

ratios, a, b, c, is given by

Page 8 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

Therefore the equation of the required line is

Question 7: . Write its vector form.

The Cartesian equation of a line is
Answer
The Cartesian equation of the line is

The given line passes through the point (5, −4, 6). The position vector of this point is

Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of vector,
It is known that the line through position vector and in the direction of the vector is
given by the equation,

This is the required equation of the given line in vector form.

Question 8:
Find the vector and the Cartesian equations of the lines that pass through the origin and
(5, −2, 3).
Answer
The required line passes through the origin. Therefore, its position vector is given by,

The direction ratios of the line through origin and (5, −2, 3) are
(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3

Page 9 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

The line is parallel to the vector given by the equation,
The equation of the line in vector form through a point with position vector and parallel
to is,

The equation of the line through the point (x1, y1, z1) and direction ratios a, b, c is given
by,
Therefore, the equation of the required line in the Cartesian form is

Question 9:
Find the vector and the Cartesian equations of the line that passes through the points (3,
−2, −5), (3, −2, 6).
Answer
Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.
Since PQ passes through P (3, −2, −5), its position vector is given by,

The direction ratios of PQ are given by,
(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11
The equation of the vector in the direction of PQ is

The equation of PQ in vector form is given by,

The equation of PQ in Cartesian form is

Page 10 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths
i.e.,

Question 10:
Find the angle between the following pairs of lines:

(i)

(ii) and

Answer
(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,

The given lines are parallel to the vectors, and ,
respectively.

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Class XII Chapter 11 – Three Dimensional Geometry Maths

(ii) The given lines are parallel to the vectors, and ,
respectively.

Question 11:
Find the angle between the following pairs of lines:
(i)

(ii)
Answer

Page 12 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

Let and be the vectors parallel to the pair of lines,
, respectively.

and

The angle, Q, between the given pair of lines is given by the relation,

(ii) Let be the vectors parallel to the given pair of lines, and

, respectively.

Page 13 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

If Q is the angle between the given pair of lines, then

Question 12:

Find the values of p so the line and
are at right angles.

Answer
The given equations can be written in the standard form as

and

The direction ratios of the lines are −3, , 2 and respectively.

Two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular to each other, if
a1a2 + b1 b2 + c1c2 = 0

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Class XII Chapter 11 – Three Dimensional Geometry Maths

Thus, the value of p is .
Question 13:

Show that the lines and are perpendicular to each other.
Answer and

The equations of the given lines are

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular to each other, if
a1a2 + b1 b2 + c1c2 = 0

∴ 7 × 1 + (−5) × 2 + 1 × 3

= 7 − 10 + 3
=0
Therefore, the given lines are perpendicular to each other.

Question 14:
Find the shortest distance between the lines

Page 15 of 58

Class XII Chapter 11 – Three Dimensional Geometry Maths

Answer
The equations of the given lines are

It is known that the shortest distance between the lines, and , is
given by,

Comparing the given equations, we obtain

Substituting all the values in equation (1), we obtain
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