Two security sensors in a multi-storey mall emit laser beams for movement detection. Three of these beams travel along the following lines: l_1:r=(λ+1)i+(3λ+2)j+(3-λ)k and l_2:r=2(μ+1)i+(3μ-1)j+(μ+1)k l_3:r=(γ+3)i+(3γ-1)j+(1-γ)k Based on the given information, answer the following questions: (i) Show that the lines l_1 and l_2 are not parallel to each other. (ii) Show that the lines l_1 and l_3 are parallel to each other. (iii)(a) Find the shortest distance between the lines l_1 and l_2. (iii)(b) Find the shortest distance between the lines l_1 and l_3.

MathsCBSE Class 12 · Sample paper, session 2026-27 · 4 marks · Case-based

CBSE Class 12 Mathematics sample paper 2026-27: question 36

From the official CBSE sample question paper for 2026-27, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedThree Dimensional Geometry

Two security sensors in a multi-storey mall emit laser beams for movement detection. Three of these beams travel along the following lines:
$l_{1} : r = (λ + 1) \hat{i} + (3 λ + 2) \hat{j} + (3 - λ) \hat{k}$ and
$l_{2} : r = 2 (μ + 1) \hat{i} + (3 μ - 1) \hat{j} + (μ + 1) \hat{k}$
$l_{3} : r = (γ + 3) \hat{i} + (3 γ - 1) \hat{j} + (1 - γ) \hat{k}$
Based on the given information, answer the following questions:

(i)
Show that the lines $l_{1}$ and $l_{2}$ are not parallel to each other.
1 mark
(ii)
Show that the lines $l_{1}$ and $l_{3}$ are parallel to each other.
1 mark
(iii)(a)
Find the shortest distance between the lines $l_{1}$ and $l_{2}$ .
2 marks
OR
(iii)(b)
Find the shortest distance between the lines $l_{1}$ and $l_{3}$ .
2 marks

Figure in the paper: Photo of a multi-storey mall interior with two sensors mounted on pillars emitting crossing red laser beams across the atrium. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Separate each line into a constant vector plus parameter × direction vector.

Hint 2

Lines are parallel exactly when their direction vectors are scalar multiples of each other.

Hint 3

For skew lines use $\frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$ ; for parallel lines use $\frac{∣ b \times ( a _{2} - a _{1} ) ∣}{∣ b ∣}$ .

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.

(i) Rewrite the lines in vector form:
- $l_{1} : r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + 3 \hat{j} - \hat{k})$ , so $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $b = \hat{i} + 3 \hat{j} - \hat{k}$ .
- $l_{2} : r = (2 \hat{i} - \hat{j} + \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$ , so $c = 2 \hat{i} - \hat{j} + \hat{k}$ and $d = 2 \hat{i} + 3 \hat{j} + \hat{k}$ .
The ratios $\frac{1}{2}, \frac{3}{3}, \frac{- 1}{1}$ are not equal, so $b \neq = k d$ for any scalar $k$ . Hence $l_{1}$ and $l_{2}$ are not parallel.
1 mark
(ii) Similarly, $l_{3} : r = (3 \hat{i} - \hat{j} + \hat{k}) + γ (\hat{i} + 3 \hat{j} - \hat{k})$ , so $e = 3 \hat{i} - \hat{j} + \hat{k}$ and $f = \hat{i} + 3 \hat{j} - \hat{k}$ . Since $f = b$ , the lines $l_{1}$ and $l_{3}$ are parallel.
1 mark
(iii)(a) Find the two vectors needed:
- $c - a = \hat{i} - 3 \hat{j} - 2 \hat{k}$ .
- $b \times d = \hat{i} 12 \hat{j} 33 \hat{k} - 1 1 = 6 \hat{i} - 3 \hat{j} - 3 \hat{k}$ .
Then $(c - a) \cdot (b \times d) = 6 + 9 + 6 = 21 \neq = 0$ , so the lines are skew.
1 mark
(iii)(a) Here $∣ b \times d ∣ = 36 + 9 + 9 = 36$ . So $S D = \frac{( c - a ) \cdot ( b \times d )}{∣ b \times d ∣} = \frac{21}{3 6} = \frac{7}{6} = \frac{7 6}{6}$ units.
1 mark
(iii)(b) The lines are parallel with common direction $b$ , so $S D = \frac{∣ b \times ( e - a ) ∣}{∣ b ∣}$ , where $e - a = 2 \hat{i} - 3 \hat{j} - 2 \hat{k}$ .
1 mark
(iii)(b) $b \times (e - a) = \hat{i} 12 \hat{j} 3 - 3 \hat{k} - 1 - 2 = - 9 \hat{i} + 0 \hat{j} - 9 \hat{k}$ , so its magnitude is $92$ . Also $∣ b ∣ = 11$ . Hence $S D = \frac{9 2}{11} = \frac{9 22}{11}$ units.
1 mark

Answer: (i) Direction vectors not proportional ⇒ not parallel; (ii) same direction vector ⇒ parallel; (iii)(a) $\frac{7 6}{6}$ units; (iii)(b) $\frac{9 22}{11}$ units

Where the marks are

Writes $l_{1}, l_{2}$ in the form $a + λ b$ and shows the direction vectors are not proportional (1 mark)
Shows the direction vectors of $l_{1}$ and $l_{3}$ are equal (1 mark)
(iii)(a) Computes $c - a$ , $b \times d$ and the scalar triple product 21 (1 mark)
(iii)(a) Shortest distance $76 /6$ units (1 mark)
(iii)(b) Uses the parallel-lines formula $∣ b \times (e - a) ∣/∣ b ∣$ (1 mark)
(iii)(b) Shortest distance $922 /11$ units (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 paperSample paper 2026-27: every question, by section.

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CBSE Class 12 Mathematics sample paper 2026-27: question 36

Hints

Step-by-step solution

Revise the chapter

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