MathsCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2025-26: question 35

From the official CBSE sample question paper for 2025-26, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerThree Dimensional Geometry

The two lines $\frac{x - 1}{3} = - y, z + 1 = 0$ and $\frac{- x}{2} = \frac{y + 1}{2} = z + 2$ intersect at a point whose y-coordinate is 1. Find the co-ordinates of their point of intersection. Find the vector equation of a line perpendicular to both the given lines and passing through this point of intersection.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Write each line in standard symmetric form; note that $- y$ means $\frac{y - 0}{- 1}$ and $\frac{- x}{2}$ means $\frac{x}{- 2}$ .

Hint 2

Take a general point on the first line and set its $y$ -coordinate equal to 1.

Hint 3

A direction perpendicular to two given directions is their cross product.

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.

Write the first line in symmetric form: $\frac{x - 1}{3} = \frac{y - 0}{- 1} = \frac{z + 1}{0} = λ$ .
A general point on it is $(3 λ + 1, - λ, - 1)$ .
½ mark
The point of intersection has $y = 1$ , so $- λ = 1 \Rightarrow λ = - 1$ .
1 mark
The point of intersection is $(3 (- 1) + 1, 1, - 1) = (- 2, 1, - 1)$ .
Check on the second line: $\frac{- ( - 2 )}{2} = 1$ , $\frac{1 + 1}{2} = 1$ and $- 1 + 2 = 1$ , so it lies on both lines.
½ mark
Direction of the first line: $m = 3 \hat{i} - \hat{j} + 0 \hat{k}$ .
1 mark
Rewrite the second line as $\frac{x}{- 2} = \frac{y + 1}{2} = \frac{z + 2}{1}$ .
Its direction is $n = - 2 \hat{i} + 2 \hat{j} + \hat{k}$ .
½ mark
A line perpendicular to both has direction $m \times n$ :
$m \times n = \hat{i} 3 - 2 \hat{j} - 1 2 \hat{k} 01 = \hat{i} (- 1 - 0) - \hat{j} (3 - 0) + \hat{k} (6 - 2) = - \hat{i} - 3 \hat{j} + 4 \hat{k} .$
1 mark
The required line passes through $(- 2, 1, - 1)$ with this direction:
$r = (- 2 \hat{i} + \hat{j} - \hat{k}) + μ (- \hat{i} - 3 \hat{j} + 4 \hat{k}) .$
½ mark

Answer: Point of intersection $(- 2, 1, - 1)$ ; line $r = (- 2 \hat{i} + \hat{j} - \hat{k}) + μ (- \hat{i} - 3 \hat{j} + 4 \hat{k})$

Where the marks are

Writes the general point $(3 λ + 1, - λ, - 1)$ of the first line (½ mark)
Uses $y = 1$ to get $λ = - 1$ (1 mark)
Obtains the point $(- 2, 1, - 1)$ (½ mark)
Direction of the first line $3 \hat{i} - \hat{j}$ (1 mark)
Direction of the second line $- 2 \hat{i} + 2 \hat{j} + \hat{k}$ (½ mark)
Computes $m \times n = - \hat{i} - 3 \hat{j} + 4 \hat{k}$ (1 mark)
States the vector equation of the required line (½ mark)

Revise the chapter

Chapter: Three Dimensional Geometry (notes, formulas, where marks are lost and every past question)
Topic: Equation of a Line in Space

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 2025-26: every question, by section.

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CBSE Class 12 Mathematics sample paper 2025-26: question 35

Hints

Step-by-step solution

Revise the chapter

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