JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Shortest distance between two skew lines in 3D

The shortest distance between the lines (x-4)/4=(y+2)/5=(z+3)/3 and (x-1)/3=(y-3)/4=(z-4)/2 is

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A3 √(6)B6 √(3)C6 √(2)D2 √(6)

Step-by-step solutionView

Correct answer

The shortest distance between the given skew lines is $3 \sqrt{6}$ .

Option analysis

Why each option works or fails

A · 3 √(6)

The student correctly applied the formula $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ . Correctly evaluate ${\vec{a}}_{2} - {\vec{a}}_{1} = (- 3, 5, 7)$ , ${\vec{b}}_{1} \times {\vec{b}}_{2} = (- 2, 1, 1)$ , their dot product $- 2 (- 3) + 1 (5) + 1 (7) = 18$ , and divide by $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{6}$ to get $18 / \sqrt{6} = 3 \sqrt{6}$ .

B · 6 √(3)

The student inverted the factors inside and outside the radical when simplifying $18 / \sqrt{6}$ , writing $6 \sqrt{3}$ instead of $3 \sqrt{6}$ . Simplify fractions with square roots systematically: $\frac{18}{\sqrt{6}} = \frac{18 \sqrt{6}}{6} = 3 \sqrt{6}$ .

C · 6 √(2)

The student made an arithmetic error in computing the norm of the cross product, taking $∣ (- 2, 1, 1) ∣$ as $\sqrt{4 + 1 + 1} = \sqrt{6}$ but simplifying $18 / \sqrt{6}$ incorrectly to $6 \sqrt{2}$ (confusing $\sqrt{6}$ with $\sqrt{3}$ ). Multiply numerator and denominator by $\sqrt{6}$ to obtain $\frac{18 \sqrt{6}}{6} = 3 \sqrt{6}$ .

D · 2 √(6)

The student flipped signs when extracting points from the symmetric equations, using ${\vec{a}}_{1} = (4, 2, 3)$ instead of $(4, - 2, - 3)$ , leading to an incorrect scalar triple product of $12$ and distance $12 / \sqrt{6} = 2 \sqrt{6}$ . The standard symmetric form is $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ , so $y + 2$ means $y_{0} = - 2$ and $z + 3$ means $z_{0} = - 3$ .

Reviewed route

Solution

StepWorking

01Given

Line 1: ${\vec{a}}_{1} = (4, - 2, - 3)$ , ${\vec{b}}_{1} = ⟨ 4, 5, 3 ⟩$ ; Line 2: ${\vec{a}}_{2} = (1, 3, 4)$ , ${\vec{b}}_{2} = ⟨ 3, 4, 2 ⟩$ .

02Goal

Find the shortest distance $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ .

03Approach

Compute ${\vec{a}}_{2} - {\vec{a}}_{1}$ , the cross product ${\vec{b}}_{1} \times {\vec{b}}_{2}$ , their dot product (which equals the scalar triple product / determinant), and divide by the magnitude of the normal vector $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣$ .

04Execute

Compute ${\vec{a}}_{2} - {\vec{a}}_{1} = (1 - 4) \hat{i} + (3 - (- 2)) \hat{j} + (4 - (- 3)) \hat{k} = - 3 \hat{i} + 5 \hat{j} + 7 \hat{k}$ . Compute ${\vec{b}}_{1} \times {\vec{b}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{array} ∣ = \hat{i} (10 - 12) - \hat{j} (8 - 9) + \hat{k} (16 - 15) = - 2 \hat{i} + \hat{j} + \hat{k}$ . Then $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{(- 2)^{2} + 1^{2} + 1^{2}} = \sqrt{6}$ . Next, $({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) = (- 3) (- 2) + (5) (1) + (7) (1) = 6 + 5 + 7 = 18$ . Thus, $d = \frac{18}{\sqrt{6}} = 3 \sqrt{6}$ .

✓Verify

Check: $18 / \sqrt{6} = 18 \sqrt{6} / 6 = 3 \sqrt{6}$ . Both ${\vec{b}}_{1}$ and ${\vec{b}}_{2}$ are non-parallel, ensuring lines are skew, and distance is positive and non-zero.

Hints that build this answer step by step

What are the position vectors ${\vec{a}}_{1}, {\vec{a}}_{2}$ of points on lines $L_{1}$ and $L_{2}$ , and their direction vectors ${\vec{b}}_{1}, {\vec{b}}_{2}$ ?
${\vec{a}}_{1} = (4, - 2, - 3), {\vec{b}}_{1} = (4, 5, 3)$ and ${\vec{a}}_{2} = (1, 3, 4), {\vec{b}}_{2} = (3, 4, 2)$
What is the cross product of the direction vectors ${\vec{b}}_{1} \times {\vec{b}}_{2}$ ?
$(- 2, 1, 1)$
What is the shortest distance $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ ?
$3 \sqrt{6}$

✓ 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: 8 September 2026

Quick checks

Students also ask

Does it matter if we take

{\vec{a}}_{1} - {\vec{a}}_{2}

instead of

{\vec{a}}_{2} - {\vec{a}}_{1}

No, because the absolute value in the numerator ensures the distance is always non-negative.

Answer

The shortest distance between the given skew lines is $3 \sqrt{6}$ .

Why each option works or fails

A · correct: 3 √(6) - The student correctly applied the formula $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ . Correctly evaluate ${\vec{a}}_{2} - {\vec{a}}_{1} = (- 3, 5, 7)$ , ${\vec{b}}_{1} \times {\vec{b}}_{2} = (- 2, 1, 1)$ , their dot product $- 2 (- 3) + 1 (5) + 1 (7) = 18$ , and divide by $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{6}$ to get $18 / \sqrt{6} = 3 \sqrt{6}$ .
B: 6 √(3) - The student inverted the factors inside and outside the radical when simplifying $18 / \sqrt{6}$ , writing $6 \sqrt{3}$ instead of $3 \sqrt{6}$ . Simplify fractions with square roots systematically: $\frac{18}{\sqrt{6}} = \frac{18 \sqrt{6}}{6} = 3 \sqrt{6}$ .
C: 6 √(2) - The student made an arithmetic error in computing the norm of the cross product, taking $∣ (- 2, 1, 1) ∣$ as $\sqrt{4 + 1 + 1} = \sqrt{6}$ but simplifying $18 / \sqrt{6}$ incorrectly to $6 \sqrt{2}$ (confusing $\sqrt{6}$ with $\sqrt{3}$ ). Multiply numerator and denominator by $\sqrt{6}$ to obtain $\frac{18 \sqrt{6}}{6} = 3 \sqrt{6}$ .
D: 2 √(6) - The student flipped signs when extracting points from the symmetric equations, using ${\vec{a}}_{1} = (4, 2, 3)$ instead of $(4, - 2, - 3)$ , leading to an incorrect scalar triple product of $12$ and distance $12 / \sqrt{6} = 2 \sqrt{6}$ . The standard symmetric form is $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ , so $y + 2$ means $y_{0} = - 2$ and $z + 3$ means $z_{0} = - 3$ .

Step-by-step solution

Given: Line 1: ${\vec{a}}_{1} = (4, - 2, - 3)$ , ${\vec{b}}_{1} = ⟨ 4, 5, 3 ⟩$ ; Line 2: ${\vec{a}}_{2} = (1, 3, 4)$ , ${\vec{b}}_{2} = ⟨ 3, 4, 2 ⟩$ .
Goal: Find the shortest distance $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ .
Approach: Compute ${\vec{a}}_{2} - {\vec{a}}_{1}$ , the cross product ${\vec{b}}_{1} \times {\vec{b}}_{2}$ , their dot product (which equals the scalar triple product / determinant), and divide by the magnitude of the normal vector $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣$ .
Execute: Compute ${\vec{a}}_{2} - {\vec{a}}_{1} = (1 - 4) \hat{i} + (3 - (- 2)) \hat{j} + (4 - (- 3)) \hat{k} = - 3 \hat{i} + 5 \hat{j} + 7 \hat{k}$ . Compute ${\vec{b}}_{1} \times {\vec{b}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{array} ∣ = \hat{i} (10 - 12) - \hat{j} (8 - 9) + \hat{k} (16 - 15) = - 2 \hat{i} + \hat{j} + \hat{k}$ . Then $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{(- 2)^{2} + 1^{2} + 1^{2}} = \sqrt{6}$ . Next, $({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) = (- 3) (- 2) + (5) (1) + (7) (1) = 6 + 5 + 7 = 18$ . Thus, $d = \frac{18}{\sqrt{6}} = 3 \sqrt{6}$ .
Verify: Check: $18 / \sqrt{6} = 18 \sqrt{6} / 6 = 3 \sqrt{6}$ . Both ${\vec{b}}_{1}$ and ${\vec{b}}_{2}$ are non-parallel, ensuring lines are skew, and distance is positive and non-zero.

Shortest distance between two skew lines in 3D

What feels right?

The shortest distance between the given skew lines is $3 \sqrt{6}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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