RD Sharma Solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.14 is provided here. The solutions are mainly designed for students to grasp the concepts quickly and have a strong command over all the subjects. To know the correct methodology to answer any question, quickly refer to RD Sharma Solutions. By referring to the answers provided by the subject-matter experts, students can excel in their examinations. Download the solutions PDF for free from the links given below.

Download PDF of RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 14

Download PDF

carouselExampleControls111

Previous



Next

Access RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 14

EXERCISE 29.14

Q1.

Solution:

Here,

(x₁, y₁, z₁) = (2, 5, 0) and (x₂, y₂, z₂) = (0, –1, 1)

Also

(a₁, b₁, c₁) = (–1, 2, 3) and (a₂, b₂, c₂) = (2, –1, 2)

We shall evaluate the numerator first.

Let

Let us simplify,

N = (–2)[(2)(2) – (–1)(3)] – (–6)[(–1)(2) – (2)(3)] + (1)[(–1)(–1) – (2)(2)]

= –2(4 + 3) + 6(–2 – 6) + (1 – 4)

= –14 – 48 – 3

= –65

Now, we shall evaluate the denominator.

Let

Q2.

Solution:

Here,

(x₁, y₁, z₁) = (–1, –1, –1) and (x₂, y₂, z₂) = (3, 5, 7)

Also

(a₁, b₁, c₁) = (7, –6, 1) and (a₂, b₂, c₂) = (1, –2, 1)

We shall evaluate the numerator first.

Let

N = (4)[(–6)(1) – (–2)(1)] – (6)[(7)(1) – (1)(1)] + (8)[(7)(–2) – (1)(–6)]

= 4(–6 + 2) – 6(7 – 1) + 8(–14 + 6)

= –16 – 36 – 64

= –116

Now, we shall evaluate the denominator.

Let

Q3.

Solution:

We need to find the shortest distance between l₁ and l₂.

The equation of a plane containing the line l₂ is given by

(3x – y – 2z + 4) + λ(2x + y + z + 1) = 0

(3 + 2λ)x + (λ – 1) y + (λ – 2) z + (4 + λ) = 0

Direction ratios of l₁ are 2, 4, 1, and those of the line containing the shortest distance are proportional to 3 + 2λ, λ – 1 and λ – 2.

We know that if two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular to each other, then a₁a₂ + b₁b₂ + c₁c₂ = 0.

(3 + 2λ)(2) + (λ – 1)(4) + (λ – 2)(1) = 0

6 + 4λ + 4λ – 4 + λ – 2 = 0

9λ = 0

λ = 0

Thus, the plane containing line l₂ is 3x – y – 2z + 4 = 0.

We have

When α = 0, (x, y, z) = (1, 3, –2)

So, the point (1, 3, –2) lies on line l₁.

Hence, the shortest distance between the two lines is the same as the distance of the perpendicular from (1, 3, –2) onto the plane 3x – y – 2z + 4 = 0.

The length of the perpendicular drawn from (x₁, y₁, z₁) to the plane Ax + By + Cz + D = 0 is given by