JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Shortest distance between two skew lines in symmetric form

Q: Can we rescale direction vectors by multiplying by integers?

Yes, because the cross product p vector × q vector and its norm rescale by the same positive constant factor, which cancels out completely in the formula for d.

The shortest distance between the lines x + 1 = 2y = -12z and x = y + 2 = 6z - 6 is

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A3/2B2C5/2D3

Step-by-step solutionView

Correct answer

The shortest distance between the two lines is 2.

Option analysis

Why each option works or fails

A · 3/2

Computing the numerator $∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣$ or the magnitude $∣ b_{1} \times b_{2} ∣$ with an arithmetic error, leading to an incorrect ratio of $\frac{3}{2}$ . Double-check vector cross products and scalar dot products carefully: the scalar triple product evaluates to 12 and the magnitude of the normal vector is 6, yielding $12 / 6 = 2$ .

B · 2

Correct option. Writing both lines in standard symmetric form gives points $a_{1} = (- 1, 0, 0)$ , $a_{2} = (0, - 2, 1)$ and direction vectors proportional to $b_{1} = ⟨ 6, 3, - 1 ⟩$ , $b_{2} = ⟨ 6, 6, 1 ⟩$ . The shortest distance formula $d = \frac{∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣}{∣ b_{1} \times b_{2} ∣}$ gives $\frac{∣ 9 - 6 + 9 ∣}{\sqrt{9 + 144 + 324} / 3} = \frac{12}{6} = 2$ .

C · 5/2

Incorrectly rewriting the standard symmetric form of $6 z - 6$ as $6 (z - 1)$ resulting in a point coordinate error, such as taking $z_{2} = - 1$ instead of $1$ . Ensure expressions like $6 z - 6$ are factored as $6 (z - 1)$ so the numerator becomes $z - 1$ , giving point coordinate $z = 1$ .

D · 3

Misidentifying the direction ratios by inverting coefficients incorrectly, such as taking coefficients directly as direction ratios instead of their reciprocals. To find direction ratios from $x + 1 = 2 y = - 12 z$ , divide by a common multiple like 12 to get $\frac{x + 1}{12} = \frac{y}{6} = \frac{z}{- 1}$ , giving direction vector $⟨ 12, 6, - 1 ⟩$ .

Reviewed route

Solution

StepWorking

01Given

Two lines: $L_{1} : x + 1 = 2 y = - 12 z$ and $L_{2} : x = y + 2 = 6 z - 6$ .

02Goal

Find the shortest distance $d$ between lines $L_{1}$ and $L_{2}$ .

03Approach

Rewrite both lines in standard symmetric Cartesian form $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ to identify base points $\vec{a}, \vec{b}$ and direction vectors $\vec{p}, \vec{q}$ . Clear fractions in direction vectors for simpler cross products, then apply $d = \frac{∣ (\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) ∣}{∣ \vec{p} \times \vec{q} ∣}$ .

04Execute

Express $L_{1}$ in standard form: $\frac{x - (- 1)}{1} = \frac{y - 0}{1 / 2} = \frac{z - 0}{- 1 / 12}$ , so point $A = (- 1, 0, 0)$ and direction vector $\vec{p} = 12 (1, \frac{1}{2}, - \frac{1}{12}) = ⟨ 12, 6, - 1 ⟩$ . Express $L_{2}$ in standard form: $\frac{x - 0}{1} = \frac{y - (- 2)}{1} = \frac{z - 1}{1 / 6}$ , so point $B = (0, - 2, 1)$ and direction vector $\vec{q} = 6 (1, 1, \frac{1}{6}) = ⟨ 6, 6, 1 ⟩$ .

05Execute

Find $\vec{b} - \vec{a} = ⟨ 0 - (- 1), - 2 - 0, 1 - 0 ⟩ = ⟨ 1, - 2, 1 ⟩$ . Compute the normal vector $\vec{n} = \vec{p} \times \vec{q}$ : $\vec{n} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & - 1 \\ 6 & 6 & 1 \end{array} ∣ = \hat{i} (6 - (- 6)) - \hat{j} (12 - (- 6)) + \hat{k} (72 - 36) = 12 \hat{i} - 18 \hat{j} + 36 \hat{k} = 6 ⟨ 2, - 3, 6 ⟩$ .

06Execute

Calculate the magnitude of $⟨ 2, - 3, 6 ⟩$ : $∣ ⟨ 2, - 3, 6 ⟩ ∣ = \sqrt{2^{2} + (- 3)^{2} + 6^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ . Compute the scalar triple product component: $(\vec{b} - \vec{a}) \cdot ⟨ 2, - 3, 6 ⟩ = 1 (2) + (- 2) (- 3) + 1 (6) = 2 + 6 + 6 = 14$ . Then $d = \frac{∣ 14 ∣}{7} = 2$ .

✓Verify

Check scalar triple product directly via determinant: $∣ \begin{array}{ccc} 1 & - 2 & 1 \\ 12 & 6 & - 1 \\ 6 & 6 & 1 \end{array} ∣ = 1 (12) - (- 2) (18) + 1 (36) = 12 + 36 + 36 = 84$ . $∣ \vec{p} \times \vec{q} ∣ = 6 \times 7 = 42$ . $d = \frac{84}{42} = 2$ . Consistent and positive.

Hints that build this answer step by step

What are the standard symmetric forms, a point, and a direction vector for each of the two lines?
Line 1: passes through $(- 1, 0, 0)$ with direction $⟨ 6, 3, - 0.5 ⟩$ ; Line 2: passes through $(0, - 2, 1)$ with direction $⟨ 6, 6, 1 ⟩$
Using direction vectors $b_{1} = ⟨ 12, 6, - 1 ⟩$ and $b_{2} = ⟨ 6, 6, 1 ⟩$ , what is the cross product $b_{1} \times b_{2}$ and its magnitude?
$b_{1} \times b_{2} = ⟨ 12, - 18, 36 ⟩$ with magnitude $\sqrt{12^{2} + (- 18)^{2} + 36^{2}} = 42$
With $a_{2} - a_{1} = ⟨ 0 - (- 1), - 2 - 0, 1 - 0 ⟩ = ⟨ 1, - 2, 1 ⟩$ , what is the shortest distance $d = \frac{∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣}{∣ b_{1} \times b_{2} ∣}$ ?
$d = \frac{∣ 1 (12) + (- 2) (- 18) + 1 (36) ∣}{42} = \frac{84}{42} = 2$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Can we rescale direction vectors by multiplying by integers?

Yes, because the cross product $\vec{p} \times \vec{q}$ and its norm rescale by the same positive constant factor, which cancels out completely in the formula for $d$ .

Answer

The shortest distance between the two lines is 2.

Why each option works or fails

A: 3/2 - Computing the numerator $∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣$ or the magnitude $∣ b_{1} \times b_{2} ∣$ with an arithmetic error, leading to an incorrect ratio of $\frac{3}{2}$ . Double-check vector cross products and scalar dot products carefully: the scalar triple product evaluates to 12 and the magnitude of the normal vector is 6, yielding $12 / 6 = 2$ .
B · correct: 2 - Correct option. Writing both lines in standard symmetric form gives points $a_{1} = (- 1, 0, 0)$ , $a_{2} = (0, - 2, 1)$ and direction vectors proportional to $b_{1} = ⟨ 6, 3, - 1 ⟩$ , $b_{2} = ⟨ 6, 6, 1 ⟩$ . The shortest distance formula $d = \frac{∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣}{∣ b_{1} \times b_{2} ∣}$ gives $\frac{∣ 9 - 6 + 9 ∣}{\sqrt{9 + 144 + 324} / 3} = \frac{12}{6} = 2$ .
C: 5/2 - Incorrectly rewriting the standard symmetric form of $6 z - 6$ as $6 (z - 1)$ resulting in a point coordinate error, such as taking $z_{2} = - 1$ instead of $1$ . Ensure expressions like $6 z - 6$ are factored as $6 (z - 1)$ so the numerator becomes $z - 1$ , giving point coordinate $z = 1$ .
D: 3 - Misidentifying the direction ratios by inverting coefficients incorrectly, such as taking coefficients directly as direction ratios instead of their reciprocals. To find direction ratios from $x + 1 = 2 y = - 12 z$ , divide by a common multiple like 12 to get $\frac{x + 1}{12} = \frac{y}{6} = \frac{z}{- 1}$ , giving direction vector $⟨ 12, 6, - 1 ⟩$ .

Step-by-step solution

Given: Two lines: $L_{1} : x + 1 = 2 y = - 12 z$ and $L_{2} : x = y + 2 = 6 z - 6$ .
Goal: Find the shortest distance $z_{2} = - 1$ between lines $L_{1}$ and $L_{2}$ .
Approach: Rewrite both lines in standard symmetric Cartesian form $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ to identify base points $\vec{a}, \vec{b}$ and direction vectors $\vec{p}, \vec{q}$ . Clear fractions in direction vectors for simpler cross products, then apply $d = \frac{∣ (\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) ∣}{∣ \vec{p} \times \vec{q} ∣}$ .
Execute: Express $L_{1}$ in standard form: $\frac{x - (- 1)}{1} = \frac{y - 0}{1 / 2} = \frac{z - 0}{- 1 / 12}$ , so point $A = (- 1, 0, 0)$ and direction vector $\vec{p} = 12 (1, \frac{1}{2}, - \frac{1}{12}) = ⟨ 12, 6, - 1 ⟩$ . Express $L_{2}$ in standard form: $\frac{x - 0}{1} = \frac{y - (- 2)}{1} = \frac{z - 1}{1 / 6}$ , so point $B = (0, - 2, 1)$ and direction vector $\vec{q} = 6 (1, 1, \frac{1}{6}) = ⟨ 6, 6, 1 ⟩$ .
Execute: Find $\vec{b} - \vec{a} = ⟨ 0 - (- 1), - 2 - 0, 1 - 0 ⟩ = ⟨ 1, - 2, 1 ⟩$ . Compute the normal vector $\vec{n} = \vec{p} \times \vec{q}$ : $\vec{n} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & - 1 \\ 6 & 6 & 1 \end{array} ∣ = \hat{i} (6 - (- 6)) - \hat{j} (12 - (- 6)) + \hat{k} (72 - 36) = 12 \hat{i} - 18 \hat{j} + 36 \hat{k} = 6 ⟨ 2, - 3, 6 ⟩$ .
Execute: Calculate the magnitude of $⟨ 2, - 3, 6 ⟩$ : $∣ ⟨ 2, - 3, 6 ⟩ ∣ = \sqrt{2^{2} + (- 3)^{2} + 6^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ . Compute the scalar triple product component: $(\vec{b} - \vec{a}) \cdot ⟨ 2, - 3, 6 ⟩ = 1 (2) + (- 2) (- 3) + 1 (6) = 2 + 6 + 6 = 14$ . Then $d = \frac{∣ 14 ∣}{7} = 2$ .
Verify: Check scalar triple product directly via determinant: $∣ \begin{array}{ccc} 1 & - 2 & 1 \\ 12 & 6 & - 1 \\ 6 & 6 & 1 \end{array} ∣ = 1 (12) - (- 2) (18) + 1 (36) = 12 + 36 + 36 = 84$ . $∣ \vec{p} \times \vec{q} ∣ = 6 \times 7 = 42$ . $d = \frac{84}{42} = 2$ . Consistent and positive.

Shortest distance between two skew lines in symmetric form

What feels right?

The shortest distance between the two lines is 2.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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