Find the coordinates of the image of the point (1,6,3) with respect to the line r=( j+2 k)+λ( i+2 j+3 k); where 'λ' is a scalar. Also, find the distance of the image from the y-axis.

MathsCBSE Class 12 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2023-24: question 35(a)

From the official CBSE sample question paper for 2023-24, section D, question 35(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(a) · 5 marks · Long answerThree Dimensional Geometry

Find the coordinates of the image of the point $(1, 6, 3)$ with respect to the line $r = (\hat{j} + 2 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k})$ ; where ' $λ$ ' is a scalar. Also, find the distance of the image from the $y$ -axis.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

First find the foot of the perpendicular from the point to the line using a general point on the line.

Hint 2

The vector from P to the foot must be perpendicular to the line's direction $(1, 2, 3)$ .

Hint 3

The foot is the mid-point of P and its image; distance of $(x, y, z)$ from the y-axis is $x^{2} + z^{2}$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

The line in Cartesian form is $\frac{x - 0}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = λ$ . Any point on it is $(λ, 2 λ + 1, 3 λ + 2)$ . Let the foot of the perpendicular from $P (1, 6, 3)$ be $L (λ, 2 λ + 1, 3 λ + 2)$ .
½ mark
Direction ratios of $P L$ : $λ - 1, 2 λ - 5, 3 λ - 1$ .
½ mark
$P L ⊥$ line (d.r. $1, 2, 3$ ): $(λ - 1) + 2 (2 λ - 5) + 3 (3 λ - 1) = 0 \Rightarrow 14 λ - 14 = 0 \Rightarrow λ = 1$ . So $L = (1, 3, 5)$ .
1 mark
Let $Q (x_{1}, y_{1}, z_{1})$ be the image of $P$ . Then $L$ is the mid-point of $P Q$ : $\frac{x _{1} + 1}{2} = 1, \frac{y _{1} + 6}{2} = 3, \frac{z _{1} + 3}{2} = 5$ .
1 mark
$x_{1} = 1, y_{1} = 0, z_{1} = 7$ . The image is $Q (1, 0, 7)$ .
1 mark
Distance of $(1, 0, 7)$ from the $y$ -axis $= x_{1}^{2} + z_{1}^{2} = 1^{2} + 7^{2} = 50 = 52$ units.
1 mark

Answer: Image $(1, 0, 7)$ ; distance from y-axis $50 = 52$ units

Where the marks are

Writes the general point $(λ, 2 λ + 1, 3 λ + 2)$ on the line (½ mark)
Finds direction ratios of $P L$ (½ mark)
Uses perpendicularity to get $λ = 1$ and foot $L (1, 3, 5)$ (1 mark)
Uses $L$ as mid-point of $P$ and its image (1 mark)
Obtains the image $(1, 0, 7)$ (1 mark)
Distance from y-axis $50$ units (1 mark)

Revise the chapter

Chapter: Three Dimensional Geometry (notes, formulas, where marks are lost and every past question)
Topic: Foot of Perpendicular and Image of a Point

Practise this chapterBoard-style questions from Three Dimensional Geometry, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Mathematics sample paper 2023-24: question 35(a)

Hints

Step-by-step solution

Revise the chapter

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