NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Maths
Chapter : 11
Chapter Name : Three Dimensional Geometry

Q1 If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, nd its direction
cosines.

Answer.
Let direction cosines of the line be l, m, and n
l = cos90 ∘ = 0
1
m = cos135 ∘ = −
√2
1
n = cos45 ∘ =
√2
1 1
Therefore, the direction cosines of the line are 0, − , and .
√2 √2

Page : 467 , Block Name : Exercise 11.1

Q2 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 a with each of the coordinate axes.
∴ l = cosa, m = cosa, n = cos
l2 + m2 + n2 = 1
⇒ cos 2α + cos 2α + cos 2α = 1
⇒ 3cos 2α = 1
1
⇒ cos 2α = 3
1
⇒ cosα = ±
√3
Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are
1 1 1
± , ± , and ± .
√3 √3 √3

Page : 467 , Block Name : Exercise 11.1

Q3 If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?

Answer. If a line has the direction ratios of -18, 12, and -4 then its direction cosines are

− 18 12 −4
, ,
√ ( − 18 )2 + ( 12 )2 + ( − 4 ) √ ( − 18 )2 + ( 12 )2 + ( − 4 ) √ ( − 18 )2 + ( 12 )2 + ( − 4 )2
2 2

− 18 12 −4
i.e... 22 , 22 ⋅ 22
−9 6 −2
, ,
11 11 11
9 6 −2
Thus, the direction cosines are − 11 , 11 , and 11 .

Page : 467 , Block Name : Exercise 11.1

Q4 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, x 1, y 1, z 1 and x 2, y 2, z 2 ( )
are given by, x 2 − x 1, y 2 − y 1 , and z 2 − z 1
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.

Page : 467 , Block Name : Exercise 11.1

Q5 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 sides AB are ( − 1 − 3), (1 − 5), and (2 − ( − 4)) L.e., − 4, − 4, and 6.
Then, ( − 4) 2 + ( − 4) 2 + (6) 2 = √16 + 16 + 36
√
= √68
= 2√17
Therefore, the direction cosines of AB are
−4 −4 6
⋅ ,
√ ( − 4 )2 + ( − 4 )2 + ( 6 )2 √ ( − 4 )2 + ( − 4 )2 + ( 6 )2 √ ( − 4 )2 + ( − 4 )2 + ( 6 )2

−4 4 6
, − ⋅
2√17 2√17 2√17

−2 −2 3
⋅ ,
√17 √17 √17
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
−4 −6 −4
+ ,
√ ( − 4 )2 + ( − 6 )2 + ( − 4 ) 2
√ ( − 4 )2 + ( − 6 )2 + ( − 4 ) √ ( − 4 )2 + ( − 6 )2 + ( − 4 )2
2

−4 −6 −4
⋅ ⋅
2√17 2√17 2√17

The direction ratios of CA are ( − 5 − 3), ( − 5 − 5), and ( − 2 − ( − 4)) l.e., − 8, − 10, and 2
Therefore, the direction cosines of AC are
−8 −5 2
, ⋅
√ ( − 8 )2 + ( 10 )2 + ( 2 ) √ ( − 8 )2 + ( 10 )2 + ( 2 )
2 2
√ ( − 8 )2 + ( 10 )2 + ( 2 )2
−8 − 10 2
, ,
2√42 2√42 2√42

Page : 467 , Block Name : Exercise 11.1

Q1 Show that the three lines with direction cosines
12 −3 −4 4 12 3 3 − 4 12
, , ; , , ; , ,
13 13 13 13 13 13 13 13 13
are mutually perpendicular.

Answer.
Two lines with direction cosines, l 1, m 1, n 1 and l 2, m 2, n 2, are perpendicular to each
other, if I 1I 2 + m 1m 2 + n 1n 2 = 0
12 −3 −4 4 12 3
(i) For the lines with direction cosines, 13 , 13 , 13 and 13 , 13 , 13 , we obtain

l 1 l 2 + m 1 m 2 + n 1n 2 =
12
13
×
4
13
+
( ) −3
13
×
12
13
+
( )−4
13
×
3
13
48 36 12
= − −
169 169 169
=0
Therefore, the lines are perpendicular.
4 12 3 3 − 4 12
(ii) For the lines with direction cosines, 13 , 13 , 13 and 13 , 13 , 13 , we obtain

l 1 l 2 + m 1 m 2 + n 1n 2 =
4
13
×
3
13
+
12
13
×
( )
−4
13
+
13
3
×
12
13
12 48 36
− + =
169 169 169
=0
Therefore, the lines are perpendicular.
Thus, all the lines are mutually perpendicular.

Page : 477 , Block Name : Exercise 11.2

Q2 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, a 1, b 1, c 1, of AB are (3 − 1), (4 − ( − 1)), and ( − 2 − 2) i.e.t 2, 5, and
−4 .
The direction ratios, a 2, b 2, c 2 of co are (3 − 0), (5 − 3), and (6 − 2) i.e., 3, 2, and 4.
AB and CD will be perpendicular to each other, if a 1a 2 + b 1b 2 + c 1c 2 = 0
a 1a 2 + b 1b 2 + c 1c 2 = 2 × 3 + 5 × 2 + ( − 4) × 4
= 6 + 10 − 16
=0
Therefore, AB and CD are perpendicular to each other.

Page : 477 , Block Name : Exercise 11.2

Q3 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) adn CD be the through the
points (-1, -2, 1) and (1, 2, 5).
The directions ratios, a 1, b 1, c 1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., − 2, − 4, and
−4 .
The direction ratios, a 2, b 2, c 2, of CD are (1 − ( − 1)), (2 − ( − 2)), and (5 − 1) l.e., 2, 4,
and 4.
a1 b1 c1
AB will be parallel to CD, if a = b = c
2 2 2
a1 −2
a2 = 2 = − 1
b1 −4
b2 = 4 = − 1
c1 −4
c2 = 4 = − 1
a1 b1 c1
∴ a = b = c
2 2 2
Thus, AB is parallel to CD.

Page : 477 , Block Name : Exercise 11.2

Q4 Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the
vector 3î + 2ĵ − 2k̂.

Answer. It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector
through A is →
a = î + 2ĵ + 3k̂
→
b = 3î + 2ĵ − 2k̂
→
It is known the the line which passes through point A and parallel to 0 b is given by
→
→
a + λ b, where λ is a constant.
r =→
⇒ →r = î + 2ĵ + 3k̂ + λ(3î + 2ĵ − 2k̂)
This is the required equation of the line.

Page : 477 , Block Name : Exercise 11.2

Q5 Find the equation of the line in vector and in cartesian form that passes through the point with
position vector
2î − j + 4k̂ and is in the direction î + 2ĵ − k̂.

Answer.
It is given that the line passes through the point with position vector
¯
a = 2î + ĵ + 4k̂
→
b = î + 2ĵ − k̂
¯
→
It is known that a line through a point with position vector a and parallel to b is given by
→
the equation, →r = →
a + λb
⇒ →r = 2î − ĵ + 4k̂ + λ(î + 2ĵ − k̂)
This is the required equation of the line in vector form.
→
r = xî − yĵ + zk̂
⇒ xî − yĵ + zk̇ = (λ + 2)î + (2λ − 1)ĵ + ( − λ + 4)k̂
Eliminating λ r we obtain the Cartesian form equation as
x−2 y+1 z−4
1 = 2 = −1

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

Page : 477 , Block Name : Exercise 11.2

Q6 Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel
to the line given by
x+3 y−4 z+8
3 = 5 = 6 .

Answer.

It is given that the line passes through the point ( − 2, 4, − 5) and is parallel to
x+3 y−4 z+8
3
= 5 = 6
x+3 y−4 z+8
3
= 5 = 6
y−4 y−4 z+8
The direction ratios of the line, 3 = 5 = 6 , are 3, 5, and 6.
x+3 y−4 z+8
The required line is parallel to 3 = 5 = 6
Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0

( )
It is known that the equation of the line through the point x 1, y 1, z 1 and with direction
x − x1 y − y1 z − z1
ratios, a, b, c, is given by a
= b
= c
Therefore the equation of the required line is
x+2 y−4 z+5
3k
= 5k = 6k
x+2 y−4 z+5
⇒ 3 = 5 = 6 =k

Page : 477 , Block Name : Exercise 11.2

x−5 y+4 z−6
Q7 The cartesian equation of a line is 3 = 7 = 2 . Write its vector form.

Answer.
The Cartesian equation of the line is
x−5 y+4 z−6
3
= 7 = 2

The given line passes through the point (5, − 4, 6) . The position vector of this point is
¯
a = 5î − 4ĵ + 6k̂
Also, the direction ratios of the given line are 3, 7 and 2.
→
This means that the line is in the direction of vector b = 3î + 7ĵ + 2k̂
→
It is known that the line through position vector b and in the direction of the vector →
a is given by
→
the equation , r = a + λ b, λ ∈ R
→ →

⇒ →r = (5î − 4ĵ + 6k̇) + λ(3î + 7ĵ + 2k̂)
This is the required equation of the given line in vector form.

Page : 477 , Block Name : Exercise 11.2

Q8 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,
→ →
a =0

The direction ratios of the line through origin and (5, -2, 3) are (5-0)=5,(-2-0)=-2,(3-0)=3
→
The line is parallel to the vector given by the equation b = 5î − 2ĵ + 3k̂
The equation of the line in vector through a point with position vector and parallel to
→ →
a + λ b, λ ∈ R
b is , →r = →
→
⇒ →r = 0 + λ(5î − 2ĵ + 3k̂)
⇒ →r = λ(5î − 2ĵ + 3k̂)

( )
The equation of the line through the point x 1, y 1, z 1 and direction ratios a, b, c is given by
x − x1 y − y1 z − z1
a
= b
= c
Therefore, the equation of the required line in the Cartesian form is
x−0 y−0 z−0
5
= −2 = 3
x y z .
⇒ 5 = −2 = 3

Page : 477 , Block Name : Exercise 11.2

Q9 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,
→
a = 3î − 2ĵ − 5k̂
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
→
b = 0î − 0, ĵ + 11k̂ = 11k̂
→
a + λ b, λ ∈ R
The equation of PQ in vector form is given by, →r = →
⇒ →r = (3î − 2ĵ − 5k̂) + 11λk̂
The equation of PQ in Cartesian form is
x − x1 y − y1 z − z1
a
= b
= c
x−3 y+2 z+5
0
= 0 = 11

Page : 478 , Block Name : Exercise 11.2

Q10 Find the angle between the following pairs of lines:
(i) →r = 2î − 5ĵ + k̂ + λ(3î + 2ĵ + 6k̂) and
→
r = 7î − 6k̂ + μ(î + 2ĵ + 2k̂)
(ii) →r = 3î + ĵ − 2k̂ + λ(î − ĵ − 2k̂) and
→
r = 2î − ĵ − 56k̂ + μ(3î − 5ĵ − 4k̂)

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

| |
→ →
b1 ⋅ b2
The angle between the given pairs of lines is given by, cosQ = → →
b1 | | b2 |

→ →
The given lines are parallel to the vectors, b 1 = 3î + 2ĵ + 6k̂ and b 2 = î + 2ĵ + 2k̂ respectively.

| | √3 2 + 2 2 + 6 2 = 7
→
∴ b1 =

| b 2 | = √(1)2 + (2)2 + (2)2 = 3
→

¯
→
b 1 ⋅ b 2 = (3î + 2ĵ + 6k̂) ⋅ (î + 2ĵ + 2k̂)
=3×1+2×2+6×2
= 3 + 4 + 12
= 19
19
⇒ cosQ = 7 × 3

⇒ Q = cos − 1 21
()
19

→ →
(ii) The given lines are parallel to the vectors, b 1 = î − ĵ − 2k̂ and b 2 = 3î − 5ĵ − 4k̂ respectively.

| | √(1)2 + ( − 1)2 + ( − 2)2 = √6
→
∴ b1 =

| b 2 | = √(3)2 + ( − 5)2 + ( − 4)2 = √50 = 5√2
→

→ →
b 1 ⋅ b 2 = (î − ĵ − 2k̂) ⋅ (3î − 5ĵ − 4k̂)
= 1 ⋅ 3 − 1( − 5) − 2( − 4)
=3+5+8
= 16
| b ⋅ b2 |
→ →

cosQ =
| b | | b2 |
→ →

16 16 16
⇒ cosQ = = =
√ 6 ⋅ 5√ 2 √ 2 ⋅ √ 3 ⋅ 5√ 2 10√3

8
⇒ cosQ =
5√ 3

⇒ Q = cos − 1
( )
5√ 3
8

Page : 478 , Block Name : Exercise 11.2

Q11 Find the angle between the following pair of lines:
x−2 y−1 z+3 x+2 y−4 z−5
(i) 2 = 5 = − 3 and − 1 = 8 = 4
x y z x−5 y−2 z−3
(ii) 2 = 2 = 1 and 4 = 1 = 8

Answer. (i)
→ →
Let b 1 and b 2 be the vectors parallel to the pair of lines,
x−2 y−1 z+3 x+2 y−4 z−5
2
= 5 = − 3 and − 1 = 8 = 4 , respectively.
→ →
∴ b 1 = 2î + 5ĵ − 3k̂ and b 2 = − î + 8ĵ + 4k̂

| b 1 | = √(2)2 + (5)2 + ( − 3)2 = √38
→

| b 2 | = √( − 1)2 + (8)2 + (4)2 = √81 = 9
→

→ →
b 1 ⋅ b 2 = (2î + 5ĵ − 3k̂) ⋅ ( − î + 8ĵ + 4k̂)
= − 2 + 40 − 12
= 26
The angle, Q, between the given pair of lines is given by the relation,

|
cosQ = b 1 ⋅ b 2 |
→ →

→
b1 b2| |
→

26
⇒ cosQ =
9√38

⇒ Q = cos − 1
( )
26
9√38

→ → x y z
(ii) Let b 1, b 2 be the vectors parallel to the given pair of lines, 2 = 2 = 1
x−5 y−5 z−3
4
= 1 = 8 , respectively.
→
b 1 = 2î + 2ĵ + k̂
→
b 2 = 4î + ĵ + 8k̂

| | √(2)2 + (2)2 + (1)2 = √9 = 3
→
∴ b1 =

| b 2 | = √42 + 12 + 82 = √81 = 9
→

→ →
b 1 ⋅ b 2 = (2î + 2ĵ + k̂) ⋅ (4î + ĵ + 8k̂)
=2×4+2×1+1×8
=8+2+8
= 18

| |
→ →
| b ⋅ b2 |
If Q is the angle between the given pair of lines, then cosQ = → →
| b | | b2 |

18 2
⇒ cosQ = 3 × 9 = 3

⇒ Q = cos − 1 3
()2

Page : 478 , Block Name : Exercise 11.2

1−x 7y − 14 z−3 7 − 7x y−5 6−z
Q12 Find the values of p so that the lines 3 = 2p
= 2 and 3p = 1 = 5 are at right
angles.

Answer. The given equations can be written in the standard form as
x−1 y−2 z−3 x−1 y−5 z−6
−3
= 2p = 2 and − 3p = 1 = − 5
7
2p − 3p
The direction ratios of the lines are − 3, 7 , 2 and 7 , 1, − 5 respectively.
Two lines with direction ratios, a 1, b 1, c 1 and a 2, b 2, c 2 are perpendicular to each other, if
a 1a 2 + b 1b 2 + c 1c 2 = 0

∴ ( − 3) ⋅ ( )()
− 3p
7
+
2p
7
⋅ (1) + 2 ⋅ ( − 5) = 0

9p 2p
⇒ 7 + 7 = 10
⇒ 11p = 70
70
⇒ p = 11
70
Thus, the value of p is 11 .

Page : 478 , Block Name : Exercise 11.2

x−5 y+2 z x y z
Q13 Show that the lines 7 = − 5 = 1 and 1 = 2 = 3 are perpendicular to each other.

Answer. The equations of the given lines are
x−5 y+2 z x y z
7
= − 5 = 1 and 1 = 2 = 3
The direction ratios of the given lines are (7,-5,1) and ( 1,2,3 ) respectively.
Two lines with direction ratios , a 1, b 1, c 1 and a 2, b 2, c 2 are perpendicular to each other, if

a 1a 2 + b 1b 2 + c 1c 2 = 0
∴ 7 × 1 + ( − 5) × 2 + 1 × 3
= 7 − 10 + 3
=0
Therefore, the given lines are perpendicular to each other.

Page : 478 , Block Name : Exercise 11.2

Q14 Find the shortest distance between the lines
→
r = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and
¯
r = 2î − ĵ − k̂ + μ(2î + ĵ + 2k̂)

Answer. The equations of the given lines are
→
r = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and
¯
r = 2î − ĵ − k̂ + μ(2î + ĵ + 2k̂)
→ →
It is known that the shortest distance between the lines, →r = → a 2 + μ b 2 is given by ,
a 1 + λ b 1 and →r = →

|
( b1 × b2 ) ⋅ ( a2 − a2 )
|
→ → → →

d=
| b1 × b2 |→ →

Comparing the given equations, we obtain
→
a 1 = î + 2ĵ + k̂
→
b 1 = î − ĵ + k̂
→
a 2 = 2î − ĵ − k̂
→
b 2 = 2î + ĵ + 2k̂
→
a2 − →
a 1 = (2î − ĵ − k̂) − (î + 2ĵ + k̂) = î − 3ĵ − 2k̂

→

→
→
b1 × b2 = 1

→
| | î

2
−1
1
ĵ k̇
1
2

b 1 × b 2 = ( − 2 − 1)î − (2 − 2)ĵ + (1 + 2)k̂ = − 3î + 3k̂

|
→
⇒ b1 × b2 =
→
| √( − 3)2 + (3)2 = √9 + 9 = √18 = 3√2
Substituting all the values in equation (1), we obtain

d=
| ( − 3î + 3k̂ ) ⋅ ( î − 3ĵ − 2k̂ )
3√ 2 |
⇒d=
| |
− 3.1 + 3 ( − 2 )
3√ 2

⇒d=
| |−9
3√ 2

3 3 × √2 3√ 2
⇒d= = = 2
√2 √2 × √2
3√ 2
Therefore, the shortest distance between the two lines is 2 units.

Page : 478 , Block Name : Exercise 11.2

Q15 Find the shortest distance between the lines
x+1 y+1 z+1 x−3 y−5 z−7
7
= − 6 = 1 and 1 = − 2 = 1

Answer. The given lines are
x+1 y+1 z+1 x−3 y−5 z−7
7
= − 6 = 1 and 1 = − 2 = 1
It is known that the shortest distance between the two lines,
x − x1 y − y1 z − z1 x − x2 y − y2 z − z2
a1
= b = c and a = b = c , is given by,
1 1 2 2 2

| |
x2 − x1 y2 − y1 z1
a1 b1 c1
a2 b2 c2

d=
√ ( bc − b c ) + ( c a − c a ) + ( a b − a b )
2 2 1
2
1 2 2 1
2
1 2 2 1
2

Comparing the given equations, we obtain
x 1 = − 1, y 1 = − 1, z 1 = − 1
a1 = 7, b 1 = − 6, c 1 = 1
x2 = 3, y 2 = 5, z 2 = 7
a2 = 1, b 2 = − 2, c 2 = 1
Then,

| || |
x2 − x1 y2 − y1 z2 − z1 4 6 8
a1 b1 c1 = 7 −6 1
a2 b2 c2 1 −2 1

= 4( − 6 + 2) − 6(7 − 1) + 8( − 14 + 6)
= − 16 − 36 − 64
= − 116
⇒ √ (b c − b c ) + (c a − c a ) + (a b − a b )
1 2 2 1
2
1 2 2 1
2
1 2 2 1
2

= √( − 6 + 2) 2 + (1 + 7) 2 + ( − 14 + 6) 2
= √16 + 36 + 64
= √116
= 2√29
Substituting distance is always non-negative, the distance between the given lines is 2√29 units.

Page : 478 , Block Name : Exercise 11.2

Q16 Find the shortest distance between the lines whose vector equations are
→
r = (î + 2ĵ + 3k̂) + λ(î − 3ĵ + 2k̂)
¯
and r = 4î + 5ĵ + 6k̂ + μ(2î + 3ĵ + k̂)

Answer. The given lines are
→
r = î + 2ĵ + 3k̂ + λ(î − 3ĵ + 2k̂) and →r = 4î + 5ĵ + 6k̂ + μ(2î + 3ĵ + k̂)
→ →
It is known that the shortest distance between the lines, →r = → a 2 + μ b 2, is given by
a 1 + λ b and →r = →

|
( b1 × b2 ) ⋅ ( a2 − a2 )
|
→ → → →

d=
| b1 × b2 | → →

→ →
Comparing the given equations with →r = → a 2 + μ b 2, we obtain
a 1 + λ b 1 and →r = →
→
a 1 = î + 2ĵ + 3k̂
→
b 1 = î − 3ĵ + 2k̂
→
a 2 = 4î + 5ĵ + 6k̂
→
b 2 = 2î + 3ĵ + k̂
→
a2 − →
a 1 = (4î + 5ĵ + 6k̇) − (î + 2ĵ + 3k̇) = 3î + 3ĵ + 3k̂

→
î
b1 × b2 = 1
2
→

| | ĵ
−3
3
k̇
2 = ( − 3 − 6)î − (1 − 4)ĵ + (3 + 6)k̂ = − 9î + 3ĵ + 9k̂
1

|
→
⇒ b1 × b2 = | √( − 9)2 + (3)2 + (9)2 = √81 + 9 + 81 = √171 = 3√19
→

( b 1 × b 2 ) ⋅ (a 2 − a 1 ) = ( − 9î + 3ĵ + 9k̂) ⋅ (3î + 3ĵ + 3k̂)
→ → → →

= −9×3+3×3+9×3
=9
Substituting all the values in equation (1), we obtain

d=
| | 3√19
9
=
3

√19
3
Therefore, the shortest distance between the two given lines is units.
√19

Page : 478 , Block Name : Exercise 11.2

Q17 Find the shortest distance between the lines whose vector equations are
→
r = (1 − t)î + (t − 2)ĵ + (3 − 2t)k̂ and
→
r = (s + 1)î + (2s − 1)ĵ − (2s + 1)k̂

Answer.
The given lines are
→
r = (1 − t)î + (t − 2)ĵ + (3 − 2t)k̂
⇒ →r = (î − 2ĵ + 3k̂) + t( − î + ĵ − 2k̂)
→
r = (s + 1)î + (2s − 1)ĵ − (2s + 1)k̂
⇒ →r = (î − ĵ + k̂) + s(î + 2ĵ − 2k̂)
→ →
It is known that the shortest distance between the lines, →r = → a 2 + μ b 2, is given by,
a 1 + λ b and →r = →

|
( b1 × b2 ) ⋅ ( a2 − a2 )
|
→ → → →

d=
| b1 × b2 |→ →

For the given equations,
→
a 1 = î − ĵ + 3k̂
→
b 1 = − î + ĵ − 2k̂
→
a 2 = î − ĵ − k̂
→
b 2 = î + 2ĵ − 2k̂
→
a2 − →
a 1 = (î − ĵ − k̂) − (i − 2ĵ + 3k̂) = ĵ − 4k̂

|
î ĵ k̇
→ →
b1 × b2 = −1 1 − 2 = ( − 2 + 4)î − (2 + 2)ĵ + ( − 2 − 1)k̂ = 2î − 4ĵ − 3k̂
1 2 −2

|→
⇒ b1 × b2 = | √(2)2 + ( − 4)2 + ( − 3)2 = √4 + 16 + 9 = √29
→

∴ ( b 1 × b 2 ) ⋅ ( a 2 − a 1 ) = (2î − 4ĵ − 3k̂) ⋅ (ĵ − 4k̂) = − 4 + 12 = 8
→ → → →

Substituting all the values in equation (3), we obtain

d=
| |
√29
8
=
8

√29
8
Therefore, the shortest distance between the lines is units.
√29

Page : 478 , Block Name : Exercise 11.2

Q1 In each of the following cases, determine the direction cosines of the normal to the plane and
the distance from the origin.
(a) z = 2 (b) x + y + z = 1
(c) 2x + 3y − z = 5 (d) 5y + 8 = 0

Answer.
(a) The equation of the plane is z = 2 or 0x + 0y + z = 2… (1)
The direction ratios of normal are 0, 0, and 1 .
∴ √0 2 + 0 2 + 1 2 = 1
Dividing both sides of equation (1) by 1, we obtain
0. x + 0. y + 1, z = 2
This is of the form / x + my + nz = d, where l, m, n are the direction cosines of normal to
the plane and d is the distance of the perpendicular drawn from the origin.
Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the
origin is 2 units.

(b) x + y + z = 1…(1)
The direction ratios of normal are 1, 1, and 1

∴ (√ (1) 2 + (1) 2 + (1) 2 = √3

Dividing both sides of equation (1) by √3, we obtain
1 1 1 1
x+ y+ z=
√3 √3 √3 √3
This equation [s of the form + my + nz = d, where l, m, n are the direction cosines of normal to the
plane and d is the distance Of normal from the origin.
1 1 1
Therefore, the direction cosines of the normal are , , and and the distance of normal form
√3 √3 √3
1
the origin is unints.
√3

(c) 2x + 3y − z = 5… (1)
The direction ratios of normal are 2, 3, and − 1 .
∴ √(2) 2 + (3) 2 + ( − 1) 2 = √14
Dividing both sides of equation (1) by √14, we obtain
2 3 1 5
x+ y− z=
√14 √14 √14 √14
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to
the plane and d is the distance of normal from the origin.
2 3 −1
Therefore, the direction cosines of the normal to the plane are , , and and the distance
√14 √14 √14
5
of normal from the origin is units.
√14

(d) 5y + 8 = 0
⇒ 0x − 5y + 0z = 8…(1)
The direction ratios of normal are 0, -5, and 0.
∴ √0 + ( − 5) 2 + 0 = 5
Dividing both sides of equation (1) by 5, we obtain
8
−y = 5
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to
the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are 0, -1, and 0 and the distance of
8
normal from the origin is 5 units.

Page : 493 , Block Name : Exercise 11.3

Q2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal
to the vector 3î + 5ȷ̂ − 6k̂.

Answer.
The normal vector is, →
n = 3î + 5ĵ − 6k̂
n→ 3î + 5ĵ − 6k̂ 3î + 5ĵ − 6k̂
∴ n̂ = | n | = → =
√ ( 3 )2 + ( 5 )2 + ( 6 ) 2 √70
It is known that the equation of the plane with position vector →r is given by, →r ⋅ n̂ = d

⇒ r̂ ⋅
( 3î + 5ĵ − 6k̂

√70 ) =7

This is the vector equation of the required plane.

Page : 493 , Block Name : Exercise 11.3

Q3 Find the Cartesian equation of the following planes:
(a) →
r ⋅ (î + ĵ − k̂) = 2 (b) →r ⋅ (2î + 3ĵ − 4k̂) = 1
(c) r ⋅ [(s − 2t)î + (3 − t)ĵ + (2s + t)k̂] = 15
→

Answer.
(a) It is given that equation of the plane is
r ⋅ (î + ĵ − k̂) = 2
→

For any arbitrary point P(x, y, z) on the plane, position vector →r is given by,
→
r = xî + yŷ − zk̂
Substituting the value of →r in equation (1), we obtain
(xî + yĵ − zk̂) ⋅ (î + ĵ − k̂) = 2
⇒x+y−z=2
This is the cartesian equation of the plane.

(b) →r ⋅ (2î + 3ĵ − 4k̂) = 1
For any arbitrary point P(x, y, z) on the plane, position vector →r is given by,
→
r = xî + yĵ − zk̂
Substituting the value of in equation (1), we obtain
(xî + yĵ + zk̂) ⋅ (2î + 3ĵ − 4k̂) = 1
⇒ 2x + 3y − 4z = 1
This is the Cartesian equation of the plane.

(c) →r ⋅ [(s − 2t)î + (3 − t)ĵ + (2s + t)k̂] = 15
For any arbitrary point P(x, y, z) on the plane, position vector →r is given by,
→
r = xî + yŷ − zk̂
Substituting the value of in equation (1), we obtain
(xî + yŷ − zk̂) ⋅ [(s − 2t)î + (3 − t)ĵ + (2s + t)k̂] = 15
⇒ (s − 2t)x + (3 − t)y + (2s + t)z = 15
This is the cartesian equation of the given plane.

Page : 493 , Block Name : Exercise 11.3

Q4 In the following cases, nd the coordinates of the foot of the perpendicular drawn from the
origin.
(a) 2x + 3y + 4z − 12 = 0 (b) 3y + 4z − 6 = 0
(c) x + y + z = 1 (d) 5y + 8 = 0

Answer. (a) Let the coordinates of the foot of perpendicular P from the origin to the plane be

( x 1, y 1, z 1 )

2x + 3y + 4z − 12 = 0
⇒ 2x + 3y + 4z = 12…(1)
The direction ratios of normal are 2, 3, and 4.
∴ √(2) 2 + (3) 2 + (4) 2 = √29
Dividing both sides of equation (1) by √29, we obtain
2 3 4 12
x+ y+ z=
√29 √29 √29 √29
This equation of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the
plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd)
Therefore, the coordinates of the foot of the perpendicular are

( √29
2
⋅
12

√29 √29
,
3
⋅
12
,
√29 √29 √29
4
,
12
i.e.. 29 , 49 , 29
) ( 24 36 48
)
(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x 1, y 1, z 1).
3y + 4z − 6 = 0
⇒ 0x + 3y + 4z = 6
The direction ratios of the normal are 0, 3, and 4.
∴ √0 + 3 2 + 4 2 = 5
Dividing both sides of equation (1) by 5, we obtain
3 4 6
0x + 5 y + 5 z = 5
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to
the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Therefore, the coordinates of the foot of the perpendicular are

( 3
0, 5 ⋅ 5 , 5 ⋅ 5
6 4 6
) ( 18 24
i.e., 0, 25 , 25 )
(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x 1, y 1, z 1).
X+y+z=1
The direction ratios of the normal are 1, 1 and 1.
∴ √1 2 + 1 2 + 1 2 = √3
Dividing both sides of equation (1) by √3 , we obtain
1 1 1 1
x+ y+ z=
√3 √3 √3 √3
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to
the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).

( 1

√3
⋅
1
,
√3 √3
1
⋅
1

√3 √3
,
1
⋅
1

√3 ) ( )
i.e.,
1 1 1
, ,
3 3 3

(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x 1, y 1, z 1).
5y + 8 = 0
⇒ 0x − 5y + 0z = 8…(1)
The direction ratios of the normal are 0, -5, and 0.
∴ √0 + ( − 5) 2 + 0 = 5
Dividing both sides of equation (1) by 5, we obtain
8
−y = 5
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to
the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd ).

( () ) (8
Therefore, the coordinates of the foot of the perpendicular are 0, − 1 5 , 0 i.e., 0, − 5 , 0
8
)
Page : 493 , Block Name : Exercise 11.3

Q5 Find the vector and cartesian equations of the planes
(a) that passes through the point (1, 0, – 2) and the normal to the plane is ˆî + ĵ − k̂.
(b) that passes through the point (1,4, 6) and the normal vector to the plane is î − 2ĵ + k̂.

Answer.
(a) The position vector of point (1, 0, − 2) is →
a = î − 2k̂
→ →
The normal vector N perpendicular to the plane is N = î + ĵ − k̂
→
The vector equation of the plane is given by, (→r − → a)N = 0
⇒ [→r − (i − 2k̂)] ⋅ (î + ĵ − k̂) = 0
r is the position vector of any point P (x, y, z) in the plane.
→

∴ →r = xî + yĵ + zk̂
Therefore, equation (1) becomes
[(xî + ŷ ĵ + zk̂) − (î − 2k̂)] ⋅ (î + ĵ − k̂) = 0
⇒ [(x − 1)î + yĵ + (z + 2)k̂] ⋅ (î + ĵ − k̂) = 0
⇒ (x − 1) + y − (z + 2) = 0
⇒x+y−z−3=0
⇒x+y−z=3
This is the Cartesian equation of the required plane.

¯
(b) The position vector of the point (1, 4, 6) is a = î + 4ĵ + 6k̂
¯
→
The normal vector N perpendicular to the plane is N = î − 2ĵ + k̂
→
The vector equation of the plane is given by, (→r − → a) ⋅ N = 0
⇒ [→r − (î + 4ĵ + 6k̂)] ⋅ (î − 2ĵ + k̂) = 0
→
( r) is the position vector of any point P (x, y, z) in the plane.

∴ →r = xî + yĵ + zk̂
Therefore, equation (1) becomes
[(xî + ŷ ĵ + zk̂) − (î + 4ĵ + 6k̂)] ⋅ (î − 2ĵ + k̂) = 0
⇒ [(x − 1)î + (y − 4)ĵ + (z − 6)k̂] ⋅ (î − 2ĵ + k̂) = 0
⇒ (x − 1) − 2(y − 4) + (z − 6) = 0
⇒ x − 2y + z + 1 = 0
This is the Cartesian equation of the required plane.

Page : 493 , Block Name : Exercise 11.3

Q6 Find the equations of the planes that passes through three points.
(a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
(b) (1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

Answer. (a) The given points are A (1, 1, – 1), B (6, 4, – 5), C (– 4, – 2, 3)

| |
1 1 −1
6 4 − 5 = (12 − 10) − (18 − 20) − ( − 12 + 16)
−4 −2 3
= 2 + 2 -4
=0
Since A, B, C are collinear points, there will be in nite number of planes passing through the given
points.
(b) The given points are A (1, 1, 0), B (1, 2, 1), C (– 2, 2, – 1)

| |
1 1 0
1 2 1 = ( − 2 − 2) − (2 + 2) = − 8 ≠ 0
−2 2 −1
Therefore, a plane will pass through the points A, B, C.

( )( )
It is known that the equation of the plane through the points , x 1, y 1, z 1 , x 2, y 2, z 2 , and

(x3, y3, z3 ), is

| |
x − x1 y − y1 z − z1
x2 − x1 y2 − y1 z2 − z1 = 0
x3 − x1 y3 − y1 z3 − z1

| |
x−1 y−1 z
⇒ 0 1 1 =0
−3 1 −1

⇒ ( − 2)(x − 1) − 3(y − 1) + 3z = 0
⇒ − 2x − 3y + 3z + 2 + 3 = 0
⇒ − 2x − 3y + 3z = − 5
⇒ 2x + 3y − 3z = 5
This is the cartesian equation of the required plane.

Page : 493 , Block Name : Exercise 11.3

Q7 Find the intercepts cut off by the plane 2x + y – z = 5.

Answer. 2x + y – z = 5
Dividing both sides of equation (1) by 5, we obtain
2 y z
5
x+ 5 − 5 =1
x y z
⇒ 5 + 5 + −5 = 1
x y z
It is known that the equation of a plane in intercept form is a + b + c = 1,
where a, b, c are the intercepts cut off by the plane at x, y and z axes respectively.
Therefore, for the given equation,
5
a = 2 , b = 5, and c = − 5
5
Thus, the intercepts cut off by the plane are 2 , 5, and − 5.

Page : 493 , Block Name : Exercise 11.3

Q8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Answer. The equation of the plane ZOX is
y=0
Any plane parallel to it is of the form, y = a
Since the y-intercept of the plane is 3,
Thus, the equation of the required plane is y =3.

Page : 493 , Block Name : Exercise 11.3

Q9 Find the equation of the plane through the intersection of the planes
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

Answer. The equation of any plane through the intersection of the planes,
3x − y + 2z − 4 = 0 and x + y + z − 2 = 0, is
(3x − y + 2z − 4) + α(x + y + z − 2) = 0, where α ∈ R
The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation (1).

∴ (3 × 2 − 2 + 2 × 1 − 4) + α(2 + 2 + 1 − 2) = 0
⇒ 2 + 3α = 0
2
⇒α= − 3
2
Substituting α = − 3 in equation (1), we obtain
2
(3x − y + 2z − 4) − 3 (x + y + z − 2) = 0

⇒ 3(3x − y + 2z − 4) − 2(x + y + z − 2) = 0
⇒ (9x − 3y + 6z − 12) − 2(x + y + z − 2) = 0
⇒ 7x − 5y + 4z − 8 = 0
This is required equation of the plane.

Page : 493 , Block Name : Exercise 11.3

Q10 Find the vector equation of the plane passing through the intersection of the planes
r ⋅ (2î + 2ĵ − 3k̂) = 7, →r ⋅ (2î + 5ĵ + 3k̂) = 9. and through the point (2, 1, 3).
→

Answer. The equations of the planes are →r ⋅ (2î + 2ĵ − 3k̂) = 7 and →r ⋅ (2î + 5ĵ + 3k̂) = 9
⇒ →r ⋅ (2î + 2ĵ − 3k) − 7 = 0 …(1)
r ⋅ (2î + 5ĵ + 3k̂) − 9
→
=0
The equation of any plane through the intersection of the planes given in equations (1)
and (2) is given by,
[→r ⋅ (2î + 2ĵ − 3k̂) − 7] + λ[→r ⋅ (2î + 5ĵ + 3k̂) − 9] = 0
r ⋅ [(2î + 2ĵ − 3k̂) + λ(2î + 5ĵ + 3k̂)] = 9λ + 7
→

r ⋅ [(2 + 2λ)î + (2 + 5λ)ĵ + (3λ − 3)k̂] = 9λ + 7
→

The line passes through the point (2, 1, 3). Therefore, its position vector is given by,
→
r = 2î + 2ĵ + 3k̂
Substituting in equation (3), we obtain
(2î + ĵ − 3k̂) ⋅ [(2 + 2λ)î + (2 + 5λ)ĵ + (3λ − 3)k̂] = 9λ + 7
⇒ (2 + 2λ) + (2 + 5λ) + (3λ − 3) = 9λ + 7
⇒ 18λ − 3 = 9λ + 7
⇒ 9λ = 10
10
⇒λ= 9

10
Substituting λ = 9 in equation (3), we obtain

→
r⋅ ( 38 68 3
9 î + 9 ĵ + 9 k̂ ) = 17

⇒ →r ⋅ (38î + 68ĵ + 3k̂) = 153
This is the vector equation of the required plane.

Page : 493 , Block Name : Exercise 11.3

Q11 Find the equation of the plane through the line of intersection of the planes
x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

Answer. The equation of the plane through the intersection of the planes, x + y + z = 1 and 2x + 3y
+ 4z = 5, is
(x + y + z − 1) + λ(2x + 3y + 4z − 5) = 0
⇒ (2λ + 1)x + (3λ + 1)y + (4λ + 1)z − (5λ + 1) = 0
The direction ratios, a 1, b 1, c 1, of this plane are (2λ + 1), (3λ + 1), and (4λ + 1)
The plane in equation (1) is perpendicular to x- y + z = 0
Its direction ratios, a 2, b 2, c 2 are 1 , -1, and 1.
Since the planes are perpendicular,
a 1a 2 + b 1b 2 + c 1c 2 = 0
⇒ (2λ + 1) − (3λ + 1) + (4λ + 1) = 0
⇒ 3λ + 1 = 0
1
⇒λ= − 3
1
Substituting λ = − 3 in equation (1), we obtain
1 1 2
3x − 3z + 3 = 0

⇒x−z+2=0
This is required equation of the plane.

Page : 493 , Block Name : Exercise 11.3

Q12 Find the angle between the planes whose vector equations are
r ⋅ (2î + 2ĵ − 3k̂) = 5 and →r ⋅ (3î − 3ĵ + 5k̂) = 3.
→

Answer. The equations of the given planes are →r ⋅ (2î + 2ĵ − 3k̂) = 5 and →r ⋅ (3î − 3ĵ + 5k̂) = 3
It is known that → n 2 are normal to the planes →r ⋅ →
n 1 and → n 2 = d 2 then the angle
n 1 = d 1 and →r ⋅ →
between them, Q, is given by,

cosQ =
|| || ||
n→ 1 ⋅ n→ 1

→
r1
→
n2

→
Here, n 1 = 2î + 2ĵ − 3k̂ and →
n 2 = 3î − 3ĵ + 5k̂
∴→ n 2 = (2î + 2ĵ − 3k̂)(3î − 3ĵ + 5k̂) = 2.3 + 2 ⋅ ( − 3) + ( − 3).5 = − 15
n1 + →

|n 1 | = √(2)2 + (2)2 + ( − 3)2 = √17
→

|n 2 | = √(3)2 + ( − 3)2 + (5)2 = √43
→

Substituting the value of n ⋅ n 2, → | | | | in equation ( 1 ) , we obtain
→ →
n 1 a and →
n2

cosQ =
| − 15

√17 ⋅ √43 |
15
⇒ cosQ =
√731

⇒ cosQ − 1 =
( ) √731
15

Page : 494 , Block Name : Exercise 11.3

Q13 In the following cases, determine whether the given planes are parallel or perpendicular, and
in case they are neither, nd the angles between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0

Answer. The direction ratios of normal to the plane, L 1 : a 1x + b 1y + c 1z = 0, are a 1, b 1, c 1 and
L 2 : a 1x + b 2y + c 2z = 0 are a 2, b 2, c 2
a1 b1 c1
L 1‖L 2, if a = b = c
2 2 2

L 1 ⊥ L 2, if a 1a 2 + b 1b 2 + c 1c 2 = 0
The angle between L 1 and L 2 is given by,
a 1a 2 + b 1b 2 + c 1c 2
Q = cos − 1 |
√a + b + c ⋅ √a + b + c
2
1
2
1
2
1
2
2
2
2
2
2

(a) The equations of the planes are 7x + 5y + 6z + 30 = 0 and 3x - y -10z + 4 = 0
Here. a t = 7. b 1 = 5. c 1 = 6
a 2 = 3, b 2 = − 1, c 2 = − 10
a 1a 2 + b 1b 2 + c 1c 2 = 7 × 3 + 5 × ( − 1) + 6 × ( − 10) = − 44 ≠ 0
Therefore, the given planes are not perpendicular.
a1 7 b1 5 c1 6 −3
a2
= 3 , b = − 1 = − 5, c = − 10 = 5
2 2
a1 b1 c1
It can be seen that, a ≠ b ≠ , c
2 2 2
Therefore, the given planes are not parallel.
The angle between them is given by,

Q = cos − 1
|√ 7 × 3 + 5 × ( − 1) + 6 × ( − 10)
(7) 2 + (5) 2 + (6) 2 × √(3) 2 + ( − 1) 2 + ( − 10) 2 |
= cos − 1
| 21 − 5 − 60
√110 × √110 |
44
= cos − 1
110
−1
2
= cos
5
(b) The equations of the planes are 2x + y + 3z − 2 = 0 and x − 2y + 5 = 0
Here, a 1 = 2, b 1 = 1, c 1 = 3 and a 2 = 1, b 2 = − 2, c 2 = 0
∴ a 1a 2 + b b + c 1c 2 = 2 × 1 + 1 × ( − 2) + 3 × 0 = 0
2

Thus, the given planes are perpendicular to each other.
(c) The equations of the given planes are 2x − 2y + 4z + 5 = 0 and 3x − 3y + 6z − 1 = 0
Here, a 1 = 2, b 1 − 2, c 1 = 4 and
a 2 = 3, b 2 = − 3, c 2 = 6a 1a 2 + b 1b 2 + c 2c 2 = 2 × 3 + ( − 2)( − 3) + 4 × 6 = 6 + 6 + 24 = 36 ≠ 0
Thus, the given planes are not perpendicular to each other.
a1 2 b1 −2 2 c1 4 2
a2
= 3 , b = − 3 = 3 and c = 6 = 3
2 2
a1 b1 c1
a2
= b = c
2 2
Thus, the given planes are parallel to each other.
(d) The equations of the planes are 2x − y + 3z − 1 = 0 and 2x − y + 3z + 3 = 0
Here, a 1 = 2, b 1 = − 1, c 1 = 3 and a 2 = 2, b 2 = − 1, c 2 = 3
a1 2 b1 −1 c1 3
a2
= 2 = 1, b = − 1 = 1 and c = 3 = 1
2 2
a1 b1 c1
a2
= b = c
2 2
Thus, the given lines are parallel to each other.
(e) The equations of the given planes are 4x + 8y + z − 8 = 0 and y + z − 4 = 0
Here, a 1 = 4, b 1 = 8, c 1 = 1 and a 2 = 0, b 2 = 1, c 2 = 1
a 1a 2 + b 1b 2 + c 1c 2 = 4 × 0 + 8 × 1 + 1 = 9 ≠ 0
Therefore, the given lines are not perpendicular to each other.
a1 4 b1 8 c1 1
a2
= 0 , b = 1 = 8, c = 1 = 1
2 2
a1 b1 c1
a2
≠ b ≠ c
2 2
Therefore, the given lines are not parallel to each other.
The angle between the planes is given by,

Q = cos − 1
| √
4×0+8×1+1×1

42 + 82 + 12 × √ 02 + 12 + 12 |
= cos − 1
| 9
9 × √2
| = cos − 1
()
1

√2
= 45 ∘

Page : 494 , Block Name : Exercise 11.3

Q14 In the following cases, nd the distance of each of the given points from the corresponding
given plane.
Point Plane
(a) (0, 0, 0) 3x − 4y + 12z = 3
(b) (3, − 2, 1) 2x − y + 2z + 3 = 0
(c) (2, 3, − 5) x + 2y − 2z = 9
(d) ( − 6, 0, 0) 2x − 3y + 6z − 2 = 0

Answer.

(
It is known that the distance between a point, p x 1, y 1, z 1 , and a plane, Ax + By + Cz = )
D, is given by,

d=
| Ax 1 + By 1 + Cz 1 − D

√A 2 + B 2 + C 2 |
(a) The given point is (0, 0, 0) and the plane is 3x − 4y + 12z = 3
3 × 0 − 4 × 0 + 12 × 0 − 3 3 3
∴d= | = = 13
√ ( 3 )2 + ( − 4 )2 + ( 12 ) 2 √169
(b) The given point is (3, − 2, 1) and the plane is 2x − y + 2z + 3 = 0

d=
| 2×3− ( −2) +2×1+3

√ ( 2 )2 + ( − 1 )2 + ( 2 ) 2
=
| 13
3 | = 3
13

(c) The given point is (2, 3, − 5) and the plane is x + 2y − 2z = 9
2+2×3−2( −5) −9 9
∴d= = 3 =3
√ ( 1 )2 + ( 2 )2 + ( − 2 ) 2

(d) The given point is ( − 6, 0, 0) and the plane is 2x − 3y + 6z − 2 = 0
2( −6) −3×0+6×0−2 − 14 14
d= | = | | = 7 =2
√ ( 2 )2 + ( − 3 )2 + ( 6 ) 2 √49

Page : 494 , Block Name : Exercise 11.3

Q1 Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line
determined by the points (3, 5, – 1), (4, 3, – 1).

Answer. Let OA be the line joining the origin, O (0, 0, 0) and the point A(2, 1, 1)
Also, let BC be the line joining the points, B (3, 5, -1) and C (4, 3, -1).

The direction ratios of OA are 2, 1, and 1 and of BC are (4 - 3) = 1, (3 - 5) = -2 and (-1 + 1) = 0
OA is perpendicular to BC, if a 1a 2 + b 1b 2 + c 1c 2 = 0
∴ a 1a 2 + b 1b 2 + c 1c 2 = 2 × 1 + 1( − 2) + 1 × 0 = 2 − 2 = 0
Thus, OA is perpendicular to BC.

Page : 497 , Block Name : Miscellaneous Exercise

Q2 If l 1, m 1, n 1 and l 2, m 2, n 2 are the direction cosines of two mutually perpendicular lines,
show that the direction cosines of the line perpendicular to both of these are
m 1n 2 − m 2n 1, n 1l 2 − n 2l 1, l 1m 2 − l 2m 1 .

Answer. It is given that l 1, m 1, n 1 and l 2, m 2, n 2 are the direction cosines of two mutually
perpendicular lines. Therefore,
l 1l 2 + m 1m 2 + n 1n 2 = 0
2 2 2
l1 + m1 + n1 = 1

l 22 + m 22 + n 22 = 1
Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction
cosines l 1, m 1 , , n 1 and l 2, m 2, n 2.
∴ ‖ 1 + mm 1 + mn 1 = 0
‖ 2 + mm 2 + m 2 = 0
l m n
∴ m n −m n = n l −n I = l m −l m
1 2 2 1 1 2 2 1 1 2 2 l
l2 m2 n2
⇒ = =
( m 1n 2 − m 2n 1 ) 2 ( n 1l 2 − n 2I 1 ) 2 ( l 1m 2 − l 2m i ) 2
l2 m2 n2
⇒ = =
( m 1n 2 − m 2n 1 ) 2 ( n 1l 2 − n 2I 1 ) 2 ( l 1m 2 − l 2m 2 ) 2
t2 + m2 + n2
=
( m 1n 2 − m 2n 1 ) 2 + ( n 1l 2 − n 2l 1 ) 2 + ( l 1m 2 − l 2m i ) 2
l, m, n are the direction cosines of the line.
∴ l 2 + m 2 + n 2 = 1…(5)
It is known that,

(l + m + n )(l + m + n ) − (l l + m m + n n )
2
1
2
1
2
1
2
2
2
2
2
2 1 2 1 2 1 2
2

( ) (
= m 1n 2 − m 2n 1 2 + n 1l 2 − n 2l 1 2 + l 1m 2 − l 2m 1 2 ) ( )
From (1), (2), and (3), we obtain

( ) ( ) (
⇒ 1.1 − 0 = m 1n 2 + m 2n 1 2 + n 1l 2 − n 2l 1 2 + l 1m 2 − l 2m 1 2 )
(m1n2 − m2n1 )2 + (n1l2 − n2I1 )2 + (l1m2 − l2m1 ) 2 = 1
Substituting the values from equations (5) and (6) in equation (4), we obtain

l2 m2 n2
= = =1
( m 1n 2 − m 2n 1 ) 2 ( n 2I 2 − n 2l 1 ) 2 ( l 1m 2 − l 2m 1 ) 2
⇒ l = m 1n 2 − m 2n 1, m = n 1l 2 − n 2I 11, n = l 1m 2 − l 2m 1
Thus, the direction cosines of the required line are m 1n 2 − m 2n 1, n 1l 2 − n 2l 1, and l 1m 2 − l 2m 1.

Page : 497 , Block Name : Miscellaneous Exercise

Q3 Find the angle between the lines whose direction ratios a, b, c and b − c, c − a, a − b.

Answer. The angle Q between the lines with direction cosines a, b, c and b-c, c-a, a-b, is given by,

cosQ =
| a(b−c) +b(c−a) +c(a−b)

√a 2 + b 2 + c 2 + √ ( b − c ) 2 + ( c − a ) 2 + ( a − b ) 2 |
⇒ Q = cos − 10
⇒ Q = 90 ∘
Thus, the angle between the lines is 90 ∘ .

Page : 498 , Block Name : Miscellaneous Exercise

Q4 Find the equation of a line parallel to x-axis and passing through the origin.

Answer. The line parallel to x-axis and passing through the origin is x-axis itself.
Let A be a point on x-axis . Therefore, the coordinates of A are given by (a, 0, 0) where a ∈ R
Direction ratios of OA are (a - 0) = a, 0, 0
The equation of OA is given by,
x−0 y−0 z−0
a
= 0 = 0
x y z
⇒ 1 = 0 = 0 =a
Thus, the equation of line parallel to x-axis and passing through origin is
x y z
1
= 0 = 0

Page : 498 , Block Name : Miscellaneous Exercise

Q5 If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2)
respectively, then nd the angle between the lines AB and CD.

Answer. The coordinates of A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively.
The direction ratios of AB are (4, -1) = 3, (5 - 2)=3, and (7 - 3) = 4
The direction ratios of CD are (2 − ( − 4)) = 6, (9 − 3) = 6, and (2 − ( − 6)) = 8
It can be seen that,
a1 b1 c1 1
a2
= b = c = 2
2 2

Therefore, AB is parallel to CD.
Thus, the angle between AB and CD is either 0 ∘ or 180 ∘ .

Page : 498 , Block Name : Miscellaneous Exercise

x−1 y−2 z−3 x−1 y−1 z−6
Q6 If the lines − 3 = 2k = 2 and 3k = 1 = − 5 are perpendicular, nd the value of k.

x−1 y−2 z−3 x−1 y−1 z−6
Answer. The direction of ratios of the lines, − 3 = 2k = 2 and 3k = 1 = − 5 are -3, 2k, 2 and
3k, 1, -5 respectively.
It is known that two lines with direction ratios, a 1, b 1, c 1 and a 2, b 2, c 2 are perpendicular, if
a 1a 2 + b 1b 2 + c 1c 2 = = 0.
∴ − 3(3k) + 2k × 1 + 2( − 5) = 0
⇒ − 9k + 2k − 10 = 0
⇒ 7k = − 10
− 10
⇒k= 7
10
Therefore, for k = − 7 , the given lines are perpendicular to each other.

Page : 498 , Block Name : Miscellaneous Exercise

Q7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane
→
r ⋅ (î + 2ĵ − 5k̂) + 9 = 0.

Answer. The position vector of the point (1, 2, 3) is →r 1 = î + 2ĵ + 3k̂
The direction ratios of the normal to the plane, →r ⋅ (î + 2ĵ − 5k̂) + 9 = 0 are 1, 2, and -5 and the
¯
normal vector is N = î + 2ĵ − 5k̂.
The equation of a line passing through a point and perpendicular to the given plane is given by,
→ →
l = →r + λN, λ ∈ R
→
⇒ l = (î + 2ĵ + 3k̂) + λ(î + 2ĵ − 5k̂)

Page : 498 , Block Name : Miscellaneous Exercise

Q8 Find the equation of the plane passing through (a, b, c) and parallel to the plane
r ⋅ (î + ĵ + k̂) = 2.
→

Answer. Any plane parallel to the plane, →r 1 ⋅ (î + ĵ + k̂) = 2 is of form →r ⋅ (î + ĵ + k̂) = λ
The plane passes through the point (a, b, c). Therefore, the position vector of this point is
→
r = aî + bĵ + ck̂
Therefore, equation (1) becomes

(aî + bĵ + ck̂) ⋅ (î + ĵ + k̂) = λ
⇒a+b+c=λ
Substituting λ = a + b + c in equation (1), we obtain
r ⋅ (î + ĵ + k̂) = a + b + c
→

This is the vector equation of the required plane.
Substituting →r = xî + yĵ + zk̂ in equation (2), we obtain
(xî + yĵ + zk̂) ⋅ (î + ĵ + k̂) = a + b + c
⇒x+y+z=a+b+c

Page : 498 , Block Name : Miscellaneous Exercise

Q9 Find the shortest distance between lines
→
r = 6î + 2ĵ + 2k̂ + λ(î − 2ĵ + 2k̂) and →r = − 4î − k̂ + μ(3î − 2ĵ − 2k̂)

Answer. The given lines are
→
r = 6î + 2ĵ + 2k̂ + λ(î − 2ĵ + 2k̂) …(1)
→
r = − 4î − k̂ + μ(3î − 2ĵ − 2k̂) …(2)
→ →
It is known that the shortest distance between two lines, →r = → a 2 + λ b 2, is given by
a 1 + λ b 1 and →r = →

|
( b1 × b2 ) ⋅ ( a2 − a1 )
|
→ → → →

d=
| b1 × b2 | → →

Comparing to equations (1) and (2), we obtain
\(

→
a 1 = 6î + 2ĵ + 2k̂
→
b 1 = î − 2ĵ + 2k̂
→
a 2 = − 4î − k̂
→
b 2 = 3î − 2ĵ − 2k̂

\)
⇒→
a2 − →
a 1 = ( − 4î − k̂) − (6î + 2ĵ + 2k̂) = − 10î − 2ĵ − 3k̂

→
⇒ b1 × b2 = 1
→

| î

3
ĵ
−2
−2 −2
k̂
2
|
= (4 + 4)î − ( − 2 − 6)ĵ + ( − 2 + 6)k̂ = 8î + 8ĵ + 4k̂

| →
∴ b1 × b2 = | √(8)2 + (8)2 + (4)2 = 12
→

( b 1 × b 2 ) ⋅ (a 2 − a 1 ) = (8î + 8ĵ + 4k̂) ⋅ ( − 10î − 2ĵ − 3k̂) = − 80 − 16 − 12 = − 108
→ → → →

Substituting all the values in equation (1) we obtain

d= | | − 108
12
=9

Therefore, the shortest distance between the two given lines is 9 units.

Page : 498 , Block Name : Miscellaneous Exercise

Q10 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-
plane.

Answer. It is known that the equation of the line passing through the points,

(x1, y1, z1 ) and (x2, y2, z2) is
x − x1 y − y1 z − z1
= =
x2 − x1 y2 − y1 z2 − z1
The line passing through the points (5, 1, 6) and (3, 4, 1) is given by,
x−5 y−1 z−6
3−5
= 4−1 = 1−6
x−5 y−1 z−6
⇒ − 2 = 3 = − 5 = k( say )
⇒ x = 5 − 2k, y = 3k + 1, z = 6 − 5k
Any point on the line is of the form (5 − 2k, 3k + 1, 6 − 5k).
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane,
5 − 2k = 0
5
⇒k= 9
5 17
⇒ 3k + 1 = 3 × 2 + 1 = 2
5 − 13
6 − 5k = 6 − 5 × 2 = 2

Therefore, the required point is 0, 2 , 2
( 17 − 13
) .

Page : 498 , Block Name : Miscellaneous Exercise

Q11 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-
plane.

( )
Answer. It is known that the equation of the line passing through the points x 1, y 1, z 1 and

(x2, y2, z2 ) is
x − x1 y − y1 z − z1
x2 − x1
= y −y = z −z
2 1 2 1
The line passing through the points (5, 1, 6) and (3, 4, 1) is given by,

x−5 y−1 z−6
3−5
= 4−1 = 1−6
x−5 y−1 z−6
⇒ − 2 = 3 = − 5 = k( say )
⇒ x = 5 − 2k, y = 3k + 1, z = 6 − 5k
Any point on the line is of the form (5 − 2k, 3k + 1, 6 − 5k).
Since the line passes through ZX-plane,
3k + 1 = 0
1
⇒k= − 3

⇒ 5 − 2k = 5 − 2 − 3 ( ) 1
= 3
17

6 − 5k = 6 − 5 − 3
( ) 1
= 3
23

Therefore, the required point is ( 17
3
23
, 0, 3 .)
Page : 498 , Block Name : Miscellaneous Exercise

Q12 Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses
the plane 2x + y + z = 7.

( )
Answer. It is known that the equation of the line passing through the points x 1, y 1, z 1 and

(x2, y2, z2 ) is
x − x1 y − y1 z − z1
x2 − x1
= y −y = z −z
2 1 2 1
The line passes through the points (3, -4, -5) and (2, -3, 1), its equation is given by,
x−3 y+4 z+5
2−3 = −3+4 = 1+5
x−3 y+4 z+5
⇒ − 1 = 1 = 6 = k( say )
⇒ x = 3 − k, y = k − 4, z = 6k − 5
Therefore, any point on the line is of the form (3 − k, k − 4, 6k − 5)
This point lies on the plane, 2x + y + z = 7
∴ 2(3 − k) + (k − 4) + (6k − 5) = 7
⇒ 5k − 3 = 7
⇒k=2
Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 × 2 − 5)i.e.,
(1, -2, 7).

Page : 498 , Block Name : Miscellaneous Exercise

Q13 Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each
of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Answer. The equation of the plane passing through the point (-1, 3, 2) is
a(x + 1) + b(y − 3) + c(z − 2) = 0…(1)
where a, b, c are the direction ratios of normal to the plane.
It is known that two planes, a 1x + b 1y + c 1z + d 1 = 0 and a 2x + b 2y + c 2z + d 2 = 0, are perpendicular,
if a 1a 2 + b 1b 2 + c 1c 2 = 0
Plane (1) is perpendicular to the plane, x + 2y + 3z = 5
∴a⋅1+b⋅2+c⋅3=0
⇒ a + 2b + 3c = 0
Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0
∴a⋅3+b⋅3+c⋅1=0
⇒ 3a + 3b + c = 0
From equations (2) and (3) we obtain
a b c
2×1−3×3
= 3×3−1×1 = 1×3−2×3
a b c
⇒ − 7 = 8 = − 3 = k( say )
⇒ a = − 7k, b = 8k, c = − 3k
Substituting the values of a, b, c in equation (1), we obtain
− 7k(x + 1) + 8k(y − 3) − 3k(z − 2) = 0
⇒ ( − 7x − 7) + (8y − 24) − 3z + 6 = 0
⇒ − 7x + 8y − 3z − 25 = 0
⇒ 7x − 8y + 3z + 25 = 0
This is the required equation of the plane.

Page : 498 , Block Name : Miscellaneous Exercise

Q14 If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane →r ⋅ (3î + 4ĵ − 12k̂) + 13 = 0
then nd the value of p.

Answer. The position vector through the point (1, 1, p ) is →
a 1 = î + ĵ + pk̂
Similarly, the position vector through the point (-3, 0, 1) is →
a 2 = − 4î + k̂
The equation of the given plane is →r ⋅ (3î + 4ĵ − 12k̂) + 13 = 0
It is known that the perpendicular distance between a point whose position vector is
¯ →
| a→ ⋅ N − d |
→
a and the plane, →r ⋅ N = d is given by D = →
|N|
¯
Here, N = 3î + 4ĵ − 12k̂ and d = − 13
Therefore, the distance between the point (1, 1, p) and the given plane is

( ( î + ĵ + pk̂ ) ⋅ ( 3î + 4ĵ − 12k̂ ) + 13
D1 = |
3î + 4ĵ − 12k̂
| 3 + 4 − 12p + 13 |
⇒ D1 =
√32 + 42 + ( − 12 )2
| 20 − 12p |
⇒ D1 = 13
Similarly, the distance between the point (-3, 0, 1) and the given plane is
| ( − 3î + k̂ ) ⋅ ( 3î + 4ĵ − 12k̂ ) + 13
D2 =
| 3î + 4ĵ − 12k̂ |
| − 9 − 12 + 13 |
⇒ D2 =
√32 + 42 + ( − 12 )2
8
⇒ D 2 = 13
It is given that the distance between the required plane and the points (1, 1, p) and (-3, 0, 1) is
equal.
∴ D1 = D2
| 20 − 12p | 8
⇒ 13
= 13
⇒ 12 − 12p = 8 or − (20 − 12p) = 8 .
⇒ 12p = 12 or 12p = 28
7
⇒ p = 1 or p = 3

Page : 498 , Block Name : Miscellaneous Exercise

Q15 Find the equation of the plane passing through the line of intersection of the planes
r ⋅ (î + ĵ + k̂) = 1 and →r ⋅ (2î + 3ĵ − k̂) + 4 = 0 and parallel to x-axis.
→

Answer. The given planes are
r ⋅ (î + ĵ + k̂) = 1
→

⇒ →r ⋅ (î + ĵ + k̂) − 1 = 0
r ⋅ (2î + 3ĵ − k̂) + 4 = 0
→

The equation of any plane passing through the line of intersection of these planes is
[→r ⋅ (î + ĵ + k̂) − 1] + λ[→r ⋅ (2î + 3ĵ − k̂) + 4] = 0
→
r ⋅ [(2λ + 1)î + (3λ + 1)ĵ + (1 − λ)k̂] + (4λ + 1) = 0
Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ)
The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis
The direction ratios of x-axis are 1, 0, and 0.
∴ 1. (2λ + 1) + 0(3λ + 1) + 0(1 − λ) = 0
⇒ 2λ + 1 = 0
1
⇒λ= − 2

1
Substituting λ = − 2 in equation (1), we obtain

⇒ →r ⋅ [ 1 3
]
− 2 ĵ + 2 k̂ + ( − 3) = 0

⇒ →r (ĵ − 3k̂) + 6 = 0
Therefore, its cartesian equation is y - 3z + 6 = 0
This is the equation of the required plane.

Page : 498 , Block Name : Miscellaneous Exercise

Q16 If O be the origin and the coordinates of P be (1, 2, – 3), then nd the equation of the plane
passing through P and perpendicular to OP.

Answer. The coordinates of the points, O and P are (0, 0, 0) and (1, 2, -3) respectively.
Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and ( − 3 − 0) = − 3
It is known that the equation of the plane passing through the point

(x1, y1, z1 ) is a (x − x1 ) + b (y − y1 ) + c (z − z1 ) = 0
where a, b, c are the direction ratios of normal.
Here, the direction ratios of normal are 1, 2, and -3 and the point P is (1, 2, -3).
Thus, the equation of the required plane is
1(x − 1) + 2(y − 2) − 3(z + 3) = 0
⇒ x + 2y − 3z − 14 = 0

Page : 498 , Block Name : Miscellaneous Exercise

Q17 Find the equation of the plane which contains the line of intersection of the planes
r ⋅ (î + 2ĵ + 3k̂) − 4 = 0, →r ⋅ (2î + ĵ − k̂) + 5 = 0 and which is perpendicular to the plane
→

r ⋅ (5î + 3ĵ − 6k̂) + 8 = 0
→

Answer. The equations of the given planes are
r ⋅ (î + 2ĵ + 3k̂) − 4 = 0
→
…(1)
r ⋅ (2î + ĵ − k̂) + 5 = 0
→
…(2)
The equation of the plane passing through the line intersection of the plane given in equation (1)
and equation (2) is
[→r ⋅ (î + 2ĵ + 3k̂) − 4] + λ[→r ⋅ (2î + ĵ − k̂) + 5] = 0
r ⋅ [(2λ + 1)î + (λ + 2)ĵ + (3 − λ)k̂] + (5λ − 4) = 0
→

The plane in equation (3) is perpendicular to the plane, →r ⋅ (5î + 3ĵ − 6k̂) + 8 = 0
∴ 5(2λ + 1) + 3(λ + 2) − 6(3 − λ) = 0
⇒ 19λ − 7 = 0
7
⇒ λ = 19
7
Substituting λ = 19 in equation (3), we obtain

⇒ →r ⋅ [ 33
19
45 50
î + 19 ĵ + 19 k̇ ] − 41
19
=0

⇒ →r ⋅ (33î + 45ĵ + 50k̂) − 41 = 0
This is the vector equation of the required plane.
The Cartesian equation of this plane can be obtained by substituting →r = xî + yĵ + zk̂ in equation
(3).
(xî + yĵ + zk̂) ⋅ (33î + 45ĵ + 50k̂) − 41 = 0
⇒ 33x + 45y + 50z − 41 = 0

Page : 498 , Block Name : Miscellaneous Exercise

Q18 Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line
r = 2î − ĵ + 2k̂ + λ(3î + 4ĵ + 2k̂) and the plane →r ⋅ (î − ĵ + k̂) = 5.
→

Answer. The equation of the given line is
→
r := 2î − ĵ + 2k̂ + λ(3î + 4ĵ + 2k̂)
The equation of the given plane is
r ⋅ (î − ĵ + k̂) = 5
→

Substituting the value of →r from equation (1) in equation (2) , we obtain
[2î − ĵ + 2k̂ + λ(3î + 4ĵ + 2k̂)] ⋅ (i − ĵ + k̂) = 5
⇒ [(3λ + 2)î + (4λ − 1)ĵ + (2λ + 2)k̂] ⋅ (î − ĵ + k̂) = 5
⇒ (3λ + 2) − (4λ − 1) + (2λ + 2) = 5
⇒λ=0
Substituting this value in equation (1), we obtain the equation of the line as
→
r = 2î − ĵ + 2k̂
This means that the position vector of the point of intersection of the line and the plane is
→
r = 2î − ĵ + 2k̂
This shows that the point of intersection of the given line and plane is given by the coordinates,
(2, − 1, 2). The point is ( − 1, − 5, − 10)
The distance d between the points (2, − 1, 2) and ( − 1, − 5, − 10) is
d= √( − 1 − 2) 2 + ( − 5 + 1) 2 + ( − 10 − 2) 2 = √9 + 16 + 144 = √169 = 13
Page : 499 , Block Name : Miscellaneous Exercise

Q19 Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes
r ⋅ (î − ĵ + 2k̂) = 5 and →r ⋅ (3î + ĵ + k̂) = 6.
→

→
Answer. Let the required line be parallel to vector b given by,
→
b = b 1î + b 2ȷ̂ + b 3k̂
The position vector of the point (1, 2, 3) is →
a = î + 2ĵ + 3k̂
The equation of line passing through (1, 2, 3) is →
a = î + 2ĵ + 3k̂

→
The equation of line passing through (1, 2, 3) and parallel to b is given by,
→ →
r =→
a + λb

(
⇒ →r (î + 2ĵ + 3k̂) + λ bî + b 2ĵ + b, k̂ )
The equations of the given planes are
r ⋅ (î − ĵ + 2k̂) = 5
→

r ⋅ (3î + ĵ + k̂) = 6
→

The line in equation (1) and plane in equation (2) are parallel.
Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

(
⇒ (î − ĵ + 2k̂) ⋅ λ b 1î + b 2ĵ + b 3k̂ = 0 )
(
⇒ λ b 1 − b 2 + 2b 3 = 0 )
⇒ b 1 − b 2 + 2b 3 = 0

(
Similarly, (3î + ĵ + k̂) ⋅ λ b 1î + b 2ĵ + b 3k̂ = 0 )
(
⇒ λ 3b 1 + b 2 + b 3 = 0 )
⇒ 3b 1 + b 2 + b 3 = 0
From equations (4) and (5), we obtain
b1 b2 b3
( −1) ×1−1×2
= 2×3−1×1 = 1×1−3( −1)
b1 b2 b3
⇒ −3 = 5 = 4
→
Therefore, the direction ratios of b are -3, 5, and 4.
→
∴ b = b 1î + b 2ĵ + b 3k̂ = − 3î + 5ĵ + 4k̂
→
Substituting the value of b in equation (1) we obtain
→
r = (î + 2ĵ + 3k̂) + λ( − 3î + 5ĵ + 4k̂)
This is the equation of the required line.

Page : 499 , Block Name : Miscellaneous Exercise

Q20 Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to
the two lines:
x−8 y + 19 z − 10 x − 15 y − 29 z−5
3
= − 16 = 7
and 3
= 8
= −5 .

→ →
Answer. Let the required line be parallel to the vector b given by, b = b 1î + b 2ĵ + b 3k̂
The position vector of the point (1, 2, -4) is →
a = î + 2ĵ − 4k̂
→
The equation of the line passing through (1, 2, -4) and parallel to vector b is
→ →
r =→a + λb

(
⇒ →r (î + 2ĵ − 4k̂) + λ b iî + b 2ĵ + b jk̂ )

The equations of the lines are
x−8 y + 19 z − 10
3
= − 16 = 7
x − 15 y − 29 z−5
3
=
8
= −5
Line (1) and line (2) are perpendicular to each other.
∴ 3b 1 − 16b 2 + 7b 3 = 0
Also, line (1) and line (3) are perpendicular to each other:
∴ 3b 1 + 8b 2 − 5b 3 = 0
From equations (4) and (5), we obtain
b1 b2 b3
( − 16 ) ( − 5 ) − 8 × 7
= 7 × 3 − 3 ( − 5 ) = 3 × 8 − 3 ( − 16 )
b1 b2 b1
⇒ 24 = 36 = 72
b1 b2 b3
⇒ 2 = 3 = 6
→
∴ Direction ratios of b are 2, 3, and 6
→
∴ b = 2î + 3ĵ + 6k̂
→
Substituting b = 2î + 3ĵ + 6k̂ in equation (1) we obtain
→
r = (î + 2ĵ − 4k̂) + λ(2î + 3ĵ + 6k̂)
This is the equation of the required line.

Page : 499 , Block Name : Miscellaneous Exercise

Q21 Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin,
1 1 1 1
then 2 + 2 + 2 =
a b c p2

Answer. The equation of a plane having intercepts a, b, c with x, y, and z axis respectively is given
by,
x y z
a
+ b + c =1
The distance (p) of the plane from the origin is given by,

p=
| √( ) ( ) ( )
1
a
0

+
0
a + b + c −1

2 1
b
0

2
+
1
c
2
|

1
⇒p⇒p=
1 1 1

√ a2
+ 2 + 2
b c

1
⇒ p2 =
1 1 1
+ 2 + 2
a2 b c
1 1 1 1
⇒ 2
= + +
p a2 b2 c2

Page : 499 , Block Name : Miscellaneous Exercise

Q22 Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
2
(A)2 units (B) 4 units (C) 8 units (D)
√29

Answer. The equations of the planes are
2x + 3y + 4z = 4
4x + 6y + 8z = 12
⇒ 2x + 3y + 4z = 6
It can be seen that the given planes are parallel.
It is known that the distance between two parallel planes
ax + by + cz = d 1 and ax + by + cz = d 2 is given by,

D=
| d2 − d1

√a 2 + b 2 + c 2 |
D=
| 6−4

√ ( 2 )2 + ( 3 )2 + ( 4 )2 |
2
D=
√29
2
Thus, the distance between the lines is units.
√29
Hence, the correct answer is D.

Page : 499 , Block Name : Miscellaneous Exercise

Q23 The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
(A) Perpendicular (B) Parallel

(C) intersect y-axis ( )
(D) passes through 0, 0, 4
5

Answer. The equations of the planes are
2x − y + 4z = 5…(1)
5x − 2.5y + 10z = 6…(2)
It can be seen that,
a1 2
=
a2 5
b1 −1 2
= =
b2 − 2.5 5
c1 4 2
= =
c2 10 5
a1 b1 c1
= =
a2 b2 c2
Therefore, the given planes are parallel.
Hence, the correct answer is B.

Page : 499 , Block Name : Miscellaneous Exercise