Find the value of p if the shortest distance between the lines r = (i + 2j + k) + λ(i - j + k) and r = (pi - j - k) + μ(2i + j + 2k) is (3)/(√(2)) units.

MathsCBSE Class 12 · Board exam 2026, set 65/5/1 · 5 marks · Long answer

CBSE Class 12 Mathematics 2026: question 35(b)

Set in the CBSE Class 12 board exam of 2026, paper 65/5/1, 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 value of p if the shortest distance between the lines

$r = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ and

$r = (p \hat{i} - \hat{j} - \hat{k}) + μ (2 \hat{i} + \hat{j} + 2 \hat{k})$

is $\frac{3}{2}$ units.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 65/5/1.

Hints

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

Hint 1

Identify $a_{1}, b_{1}, a_{2}, b_{2}$ from the two vector equations.

Hint 2

Use $d = \frac{( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} )}{∣ b _{1} \times b _{2} ∣}$ .

Hint 3

Because of the modulus, the equation in p gives two values.

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.

Here $a_{1} = \hat{i} + 2 \hat{j} + \hat{k}$ , $b_{1} = \hat{i} - \hat{j} + \hat{k}$ , $a_{2} = p \hat{i} - \hat{j} - \hat{k}$ and $b_{2} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ .
So $a_{2} - a_{1} = (p - 1) \hat{i} - 3 \hat{j} - 2 \hat{k}$ .
1 mark
$b_{1} \times b_{2} = \hat{i} 12 \hat{j} - 1 1 \hat{k} 12 = (- 2 - 1) \hat{i} - (2 - 2) \hat{j} + (1 + 2) \hat{k} = - 3 \hat{i} + 3 \hat{k}$ .
Hence $∣ b_{1} \times b_{2} ∣ = 9 + 9 = 32$ .
1½ marks
Shortest distance $d = \frac{( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} )}{∣ b _{1} \times b _{2} ∣} = \frac{- 3 ( p - 1 ) + 0 - 6}{3 2} = \frac{- 3 p - 3}{3 2}$ .
Setting this equal to the given distance: $\frac{- 3 p - 3}{3 2} = \pm \frac{3}{2}$ .
1½ marks
This gives $- 3 p - 3 = \pm 9$ , i.e. $∣ p + 1∣ = 3$ .
Therefore $p = 2$ or $p = - 4$ .
1 mark

Answer: $p = 2$ or $p = - 4$

Where the marks are

Identifies the vectors and finds $a_{2} - a_{1} = (p - 1) \hat{i} - 3 \hat{j} - 2 \hat{k}$ (1 mark)
Computes $b_{1} \times b_{2} = - 3 \hat{i} + 3 \hat{k}$ with modulus $32$ (1½ marks)
Applies the shortest distance formula and equates it to $\frac{3}{2}$ (1½ marks)
Solves to get $p = 2, - 4$ (1 mark)

Revise the chapter

Chapter: Three Dimensional Geometry (notes, formulas, where marks are lost and every past question)
Topic: Shortest Distance Between Two Lines

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 paper2026 board paper, set 65/5/1: every question, by section.

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CBSE Class 12 Mathematics 2026: question 35(b)

Hints

Step-by-step solution

Revise the chapter

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