RD Sharma Solutions Class 12 The Plane Exercise 29.15

RD Sharma Solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.15, is provided here for students to prepare for their exams with ease. Students can refer to RD Sharma Solutions for Class 12, which is the best reference material developed by our expert tutors, to help students gain knowledge and to have a good grip on the subject. Solving these exercises will ensure students obtain good scores in their exams. The PDF of this chapter’s solutions are available for free and can be downloaded using the links given below.

RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 15

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Access RD Sharma Solutions For Class 12 Maths Chapter 29 Exercise 15

EXERCISE 29.15

Q1.

Solution:

Given:

Let point P = (0, 0, 0), and M be the image of P in the plane

3x + 4y – 6z + 1 = 0.

Direction ratios of PM are proportional to 3, 4, – 6 as PM is normal to the plane.

The equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Let M = (3α, 4α, –6α).

As M is the image of P in the given plane, the midpoint of PM lies on the plane.

Using the midpoint formula, we have

Q2.

Solution:

Given:

Let point P = (1, 2, –1) and M be the image of P in the plane 3x – 5y + 4z = 5.

Direction ratios of PM are proportional to 3, –5, 4 as PM is normal to the plane.

The equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Let M = (3α + 1, –5α + 2, 4α – 1).

As M is the image of P in the given plane, the midpoint of PM lies on the plane.

Using the midpoint formula, we have

Q3.

Solution:

So, Q = (2α – 1, 3α + 3, –α + 1)

Now, let us find the direction ratios of PQ.

The direction ratios of a line joining two points (x₁, y₁, z₁) and (x₂, y₂, z₂) are given by (x₂ – x₁, y₂ – y₁, z₂ – z₁).

Here,

(x₁, y₁, z₁) = (5, 4, 2) and (x₂, y₂, z₂) = (2α – 1, 3α + 3, –α + 1)

Direction Ratios of PQ are ((2α – 1) – (5), (3α + 3) – (4), (–α + 1) – (2))

Direction Ratios of PQ are (2α – 6, 3α – 1, –α – 1)

PQ is perpendicular to the given line, whose direction ratios are (2, 3, –1).

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.

(2)(2α – 6) + (3)(3α – 1) + (–1)(–α – 1) = 0

4α – 12 + 9α – 3 + α + 1 = 0

14α – 14 = 0

14α = 14

α = 1

We have Q = (2α – 1, 3α + 3, –α + 1)

Q = (2×1 – 1, 3×1 + 3, –1 + 1)

Q = (1, 6, 0)

Using the distance formula, we have

Q4.

Solution:

Q5.

Solution:

Given:

Let point P = (1, 1, 2) and Q be the foot of the perpendicular drawn from P to the plane 2x – 2y + 4z + 5 = 0.

Direction ratios of PQ are proportional to 2, –2, 4 as PQ is normal to the plane.

The equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

This point lies on the given plane, which means this point satisfies the plane equation.

2(2α + 1) – 2(–2α + 1) + 4(4α + 2) + 5 = 0

4α + 2 + 4α – 2 + 16α + 8 + 5 = 0

24α + 13 = 0

24α = –13