MathsCBSE Class 12 · Board exam 2023, set 65/1/1 · 5 marks · Long answer

CBSE Class 12 Mathematics 2023: question 34(a)

Set in the CBSE Class 12 board exam of 2023, paper 65/1/1, section D, question 34(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(a) · 5 marks · Long answerThree Dimensional Geometry

Find the vector and the Cartesian equations of a line passing through the point $(1, 2, - 4)$ and parallel to the line joining the points $A (3, 3, - 5)$ and $B (1, 0, - 11)$ . Hence, find the distance between the two lines.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 65/1/1.

Hints

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

Hint 1

The direction ratios of the line come from the vector $A B$ .

Hint 2

Both lines have the same direction, so they are parallel. Which distance formula applies to parallel lines?

Hint 3

Use $d = ∣ (a_{2} - a_{1}) \times b ∣/∣ b ∣$ , taking $a_{1}$ as the given point and $a_{2}$ as A.

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.

Direction of $A B$ : $A B = (1 - 3) \hat{i} + (0 - 3) \hat{j} + (- 11 + 5) \hat{k} = - 2 \hat{i} - 3 \hat{j} - 6 \hat{k}$ , so we may take $b = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ as the direction. The required line passes through $(1, 2, - 4)$ , so its vector equation is $r = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ .
1 mark
Cartesian equation: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$ .
1 mark
Line $A B$ through $A (3, 3, - 5)$ with the same direction: $r = (3 \hat{i} + 3 \hat{j} - 5 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ . The two lines are parallel, so the distance between them is $d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}$ .
½ mark
Here $a_{1} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $a_{2} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k}$ , so $a_{2} - a_{1} = 2 \hat{i} + \hat{j} - \hat{k}$ .
½ mark
$(a_{2} - a_{1}) \times b = \hat{i} 22 \hat{j} 13 \hat{k} - 1 6 = \hat{i} (6 + 3) - \hat{j} (12 + 2) + \hat{k} (6 - 2) = 9 \hat{i} - 14 \hat{j} + 4 \hat{k}$ .
1 mark
$∣ (a_{2} - a_{1}) \times b ∣ = 81 + 196 + 16 = 293$ and $∣ b ∣ = 4 + 9 + 36 = 7$ . Therefore $d = \frac{293}{7}$ units.
1 mark

Answer: $r = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ ; $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$ ; distance $= \frac{293}{7}$ units

Where the marks are

Correct vector equation of the required line (1 mark)
Correct Cartesian equation (1 mark)
Equation of line AB and the parallel-lines distance formula (½ mark)
Finds $a_{2} - a_{1} = 2 \hat{i} + \hat{j} - \hat{k}$ (½ mark)
Cross product $9 \hat{i} - 14 \hat{j} + 4 \hat{k}$ (1 mark)
Distance $293 /7$ (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 paper2023 board paper, set 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2023: question 34(a)

Hints

Step-by-step solution

Revise the chapter

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