Find the image of the point (1,2,1) with respect to the line (x-3)/(1)=(y+1)/(2)=(z-1)/(3). Also find the equation of the line joining the given point and its image.

Image (39/7,-20/7,19/7 ); line: (x-1)/(16)=(y-2)/(-17)=(z-1)/(6)

MathsCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2024-25: question 35(b)

From the official CBSE sample question paper for 2024-25, section D, question 35(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

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

Find the image of the point $(1, 2, 1)$ with respect to the line $\frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 1}{3}$ . Also find the equation of the line joining the given point and its image.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

The image $Q$ is such that the foot of the perpendicular $L$ is the mid-point of $P Q$ .

Hint 2

Write a general point of the line in terms of $λ$ , then make $P L$ perpendicular to the direction $(1, 2, 3)$ .

Hint 3

Once you have $L$ , use $Q = 2 L - P$ . The line $P Q$ then passes through $P$ with direction $Q - P$ .

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.

Let $P (1, 2, 1)$ be the given point and $L$ the foot of the perpendicular from $P$ to the line. Put $\frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 1}{3} = λ$ . Any point on the line is $(λ + 3, 2 λ - 1, 3 λ + 1)$ , so take $L$ as this point.
½ mark
Direction ratios of $P L$ are $(λ + 3 - 1, 2 λ - 1 - 2, 3 λ + 1 - 1) = (λ + 2, 2 λ - 3, 3 λ)$ .
½ mark
$P L$ is perpendicular to the line, whose direction ratios are $1, 2, 3$ . So $(λ + 2) (1) + (2 λ - 3) (2) + 3 λ (3) = 0$ , which gives $14 λ - 4 = 0$ and $λ = \frac{2}{7}$ .
½ mark
Substitute $λ = \frac{2}{7}$ : the foot is $L (\frac{23}{7}, - \frac{3}{7}, \frac{13}{7})$ .
½ mark
Let $Q (x_{1}, y_{1}, z_{1})$ be the image. Then $L$ is the mid-point of $P Q$ : $\frac{1 + x _{1}}{2} = \frac{23}{7}$ , $\frac{2 + y _{1}}{2} = - \frac{3}{7}$ , $\frac{1 + z _{1}}{2} = \frac{13}{7}$ .
1 mark
Solve these: $x_{1} = \frac{39}{7}$ , $y_{1} = - \frac{20}{7}$ , $z_{1} = \frac{19}{7}$ . So the image is $Q (\frac{39}{7}, - \frac{20}{7}, \frac{19}{7})$ .
1 mark
Direction ratios of $P Q$ are $(\frac{32}{7}, - \frac{34}{7}, \frac{12}{7})$ , which are proportional to $(16, - 17, 6)$ . The line joining $P$ and $Q$ is $\frac{x - 1}{16} = \frac{y - 2}{- 17} = \frac{z - 1}{6}$ .
1 mark

Answer: Image $(\frac{39}{7}, - \frac{20}{7}, \frac{19}{7})$ ; line: $\frac{x - 1}{16} = \frac{y - 2}{- 17} = \frac{z - 1}{6}$

Where the marks are

Writes the general point $(λ + 3, 2 λ - 1, 3 λ + 1)$ for the foot $L$ (1/2 mark)
Finds the direction ratios of $P L$ (1/2 mark)
Uses the perpendicularity condition to get $λ = 2/7$ (1/2 mark)
Finds the foot $L (23/7, - 3/7, 13/7)$ (1/2 mark)
Uses $L$ as the mid-point of $P Q$ (1 mark)
Finds the image $Q (39/7, - 20/7, 19/7)$ (1 mark)
Writes the equation of line $P Q$ (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 2024-25: every question, by section.

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CBSE Class 12 Mathematics sample paper 2024-25: question 35(b)

Hints

Step-by-step solution

Revise the chapter

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