NCERT Solutions For Class 12 Maths Chapter 11 Three Dimensional Geometry

Solution:The direction cosines of a line are given by:

l = cos α,   m = cos β,   n = cos γ

Here,

• α = 90°
• β = 135°
• γ = 45°

Now calculating each cosine:

• l = cos 90° = 0
• m = cos 135° = −1/√2
• n = cos 45° = 1/√2

Therefore, the direction cosines of the line are:

(0, −1/√2, 1/√2)

Solution: The given line passes through the origin O(0, 0, 0) and the point A(5, −2, 3).The direction ratios of the line are:

⟨5, −2, 3⟩

Vector Equation of the Line:

r = t(5i − 2j + 3k), where t is a scalar.

Cartesian Equation of the Line:

The Cartesian form of the line is given by:

x/5 = y/−2 = z/3

Solution:The given lines are:

Line 1: r = 6i + 2j + 2k + λ(i − 2j + 2k)

Line 2: r = −4i − k + μ(3i − 2j − 2k)

A point on Line 1 is A(6, 2, 2) and its direction ratios are:

d₁ = ⟨1, −2, 2⟩

A point on Line 2 is B(−4, 0, −1) and its direction ratios are:

d₂ = ⟨3, −2, −2⟩

The vector AB is:

AB = ⟨−10, −2, −3⟩

The shortest distance between two skew lines is given by:

|AB · (d₁ × d₂)| / |d₁ × d₂|

First, compute the cross product:

d₁ × d₂ = ⟨8, 8, 4⟩

Now,

|AB · (d₁ × d₂)| = |(−10)(8) + (−2)(8) + (−3)(4)| = 108

|d₁ × d₂| = √(8² + 8² + 4²) = √144 = 12

Shortest distance = 108 / 12 = 9 units

Solution: For a plane of the form r · n = d, the vector n is the normal to the plane.The angle between two planes is the angle between their normals.

Normal to plane 1:

n₁ = ⟨2, 2, −1⟩

Normal to plane 2:

n₂ = ⟨3, −3, 5⟩

Angle θ between planes:

cos θ = |n₁ · n₂| / (|n₁| |n₂|)

n₁ · n₂ = (2)(3) + (2)(−3) + (−1)(5) = 6 − 6 − 5 = −5

|n₁ · n₂| = 5

|n₁| = √(2² + 2² + (−1)²) = √(4 + 4 + 1) = √9 = 3

|n₂| = √(3² + (−3)² + 5²) = √(9 + 9 + 25) = √43

Therefore,

cos θ = 5 / (3√43)

Angle between the planes:

θ = cos⁻¹( 5 / (3√43) )

Solution: Write each line in parametric form.For Line 1, let (x + 1)/7 = (y + 1)/(-6) = (z - 1)/1 = t.

So a point on Line 1 is A(−1, −1, 1) and its direction ratios are:

d₁ = ⟨7, −6, 1⟩

For Line 2, let (x − 3)/1 = (y − 5)/(-2) = (z − 7)/1 = s.

So a point on Line 2 is B(3, 5, 7) and its direction ratios are:

d₂ = ⟨1, −2, 1⟩

Vector AB = B − A = ⟨3 − (−1), 5 − (−1), 7 − 1⟩ = ⟨4, 6, 6⟩

Shortest distance between two skew lines:

|AB · (d₁ × d₂)| / |d₁ × d₂|

Compute d₁ × d₂:

d₁ × d₂ = ⟨−4, −6, −8⟩

|d₁ × d₂| = √( (−4)² + (−6)² + (−8)² ) = √(16 + 36 + 64) = √116 = 2√29

AB · (d₁ × d₂) = ⟨4, 6, 6⟩ · ⟨−4, −6, −8⟩

= (4)(−4) + (6)(−6) + (6)(−8)

= −16 − 36 − 48 = −100

|AB · (d₁ × d₂)| = 100

Therefore, shortest distance =

100 / (2√29) = 50/√29

Shortest distance = 50/√29 units

(or 50√29/29 units)

Solution: For a line in symmetric form, the denominators give the direction ratios (d.r.).For Line 1:

(x − 5)/7 = (y + 2)/(−5) = z/1

So direction ratios are d₁ = ⟨7, −5, 1⟩

For Line 2:

x/1 = y/2 = z/3

So direction ratios are d₂ = ⟨1, 2, 3⟩

Two lines are perpendicular if their direction vectors are perpendicular, i.e.

d₁ · d₂ = 0

d₁ · d₂ = (7)(1) + (−5)(2) + (1)(3)

= 7 − 10 + 3

= 0

Since d₁ · d₂ = 0, the given lines are perpendicular to each other.

Solution: Two lines are at right angles if their direction ratios are perpendicular, i.e. their dot product is 0.Line 1:

(1 − x)/3 = (7y − 14)/(2p) = (z − 3)/2 = t

1 − x = 3t ⇒ x = 1 − 3t

7y − 14 = 2pt ⇒ y = 2 + (2p/7)t

z − 3 = 2t ⇒ z = 3 + 2t

Direction ratios of Line 1: d₁ = ⟨−3, 2p/7, 2⟩

Multiply by 7 to avoid fraction: d₁ = ⟨−21, 2p, 14⟩

Line 2:

(7 − 7x)/(3p) = (y − 5)/1 = (6 − z)/5 = s

7 − 7x = 3ps ⇒ x = 1 − (3p/7)s

y − 5 = s ⇒ y = 5 + s

6 − z = 5s ⇒ z = 6 − 5s

Direction ratios of Line 2: d₂ = ⟨−3p/7, 1, −5⟩

Multiply by 7: d₂ = ⟨−3p, 7, −35⟩

For right angles:

d₁ · d₂ = 0

⟨−21, 2p, 14⟩ · ⟨−3p, 7, −35⟩ = 0

(−21)(−3p) + (2p)(7) + (14)(−35) = 0

63p + 14p − 490 = 0

77p = 490

p = 490/77 = 70/11

Hence, p = 70/11.