Find the shortest distance between the lines l_1 and l_2 whose vector equations are r=(-i-j-k)+λ(7i-6j+k) and r=(3i+5j+7k)+μ(i-2j+k) where λ and μ are parameters.

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

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

From the official CBSE sample question paper for 2024-25, 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 shortest distance between the lines $l_{1}$ and $l_{2}$ whose vector equations are
$r = (- \hat{i} - \hat{j} - \hat{k}) + λ (7 \hat{i} - 6 \hat{j} + \hat{k})$ and $r = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) + μ (\hat{i} - 2 \hat{j} + \hat{k})$
where $λ$ and $μ$ are parameters.

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

First check whether the lines are parallel by comparing their direction vectors.

Hint 2

For skew lines, use $d = \frac{( b _{1} \times b _{2} ) \cdot ( a _{2} - a _{1} )}{∣ b _{1} \times b _{2} ∣}$ .

Hint 3

Work out $b_{1} \times b_{2}$ as a $3 \times 3$ determinant, then take its dot product with $a_{2} - a_{1}$ .

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.

Compare each line with $r = a + λ b$ : $a_{1} = - \hat{i} - \hat{j} - \hat{k}$ , $b_{1} = 7 \hat{i} - 6 \hat{j} + \hat{k}$ , $a_{2} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k}$ , $b_{2} = \hat{i} - 2 \hat{j} + \hat{k}$ . Since $b_{1}$ is not a multiple of $b_{2}$ , the lines are not parallel.
1 mark
Find the vector joining the given points: $a_{2} - a_{1} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k}$ .
½ mark
Find the cross product: $b_{1} \times b_{2} = \hat{i} 71 \hat{j} - 6 - 2 \hat{k} 11 = \hat{i} (- 6 + 2) - \hat{j} (7 - 1) + \hat{k} (- 14 + 6) = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k}$ .
1½ marks
Find its magnitude: $∣ b_{1} \times b_{2} ∣ = 16 + 36 + 64 = 116$ .
½ mark
Find the dot product: $(b_{1} \times b_{2}) \cdot (a_{2} - a_{1}) = - 16 - 36 - 64 = - 116$ .
½ mark
Apply the shortest distance formula: $d = \frac{( b _{1} \times b _{2} ) \cdot ( a _{2} - a _{1} )}{∣ b _{1} \times b _{2} ∣} = \frac{116}{116} = 116 = 229$ units.
1 mark

Answer: Shortest distance $= 229$ units

Where the marks are

Identifies $a_{1}, b_{1}, a_{2}, b_{2}$ and notes the lines are non-parallel (1 mark)
Finds $a_{2} - a_{1} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k}$ (1/2 mark)
Computes $b_{1} \times b_{2} = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k}$ (1½ marks)
Finds $∣ b_{1} \times b_{2} ∣ = 116$ (1/2 mark)
Finds the dot product $= - 116$ (1/2 mark)
Gets the shortest distance $229$ 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 2024-25: every question, by section.

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

Hints

Step-by-step solution

Revise the chapter

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