JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Shortest distance between two skew lines in 3D space

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

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A5√(3)B7√(3)C6√(3)D4√(3)

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A · 5√(3)

Evaluating the scalar triple product numerator as $30 \sqrt{3}$ or miscalculating the dot product with the cross product vector leads to an incorrect coefficient of 5. Carefully compute $({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) = - 8 (2) - 7 (2) + 3 (2) = - 36$ , giving absolute value $36$ , which divides by $2 \sqrt{3}$ to give $18 / \sqrt{3} = 6 \sqrt{3}$ .

B · 7√(3)

Making an arithmetic slip when computing the difference vector ${\vec{a}}_{2} - {\vec{a}}_{1}$ or its dot product with the normal vector leads to a numerator of $42$ . Check the coordinates of the reference points: ${\vec{a}}_{1} = (5, 2, 4)$ and ${\vec{a}}_{2} = (- 3, - 5, 1)$ , giving ${\vec{a}}_{2} - {\vec{a}}_{1} = (- 8, - 7, - 3)$ and a numerator magnitude of $36$ , not $42$ .

C · 6√(3)

The student correctly applied the formula $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ . This option is correct.

D · 4√(3)

Miscalculating the cross product magnitude as $\sqrt{27}$ or making an error in simplifying $36 / \sqrt{12}$ leads to $4 \sqrt{3}$ . Evaluate $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{2^{2} + 2^{2} + 2^{2}} = \sqrt{12} = 2 \sqrt{3}$ . Dividing $36$ by $2 \sqrt{3}$ gives $18 / \sqrt{3} = 6 \sqrt{3}$ .

Reviewed route

Solution

StepWorking

01Given

Line $L_{1}$ : point ${\vec{a}}_{1} = (5, 2, 4)$ , direction vector ${\vec{b}}_{1} = \hat{i} + 2 \hat{j} - 3 \hat{k}$ . Line $L_{2}$ : point ${\vec{a}}_{2} = (- 3, - 5, 1)$ , direction vector ${\vec{b}}_{2} = \hat{i} + 4 \hat{j} - 5 \hat{k}$ .

02Goal

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

03Approach

Compute ${\vec{a}}_{2} - {\vec{a}}_{1}$ and ${\vec{b}}_{1} \times {\vec{b}}_{2}$ , then compute their dot product and divide by the magnitude $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣$ .

04Execute

Find the vector connecting the points: ${\vec{a}}_{2} - {\vec{a}}_{1} = (- 3 - 5) \hat{i} + (- 5 - 2) \hat{j} + (1 - 4) \hat{k} = - 8 \hat{i} - 7 \hat{j} - 3 \hat{k}$ .

05Execute

Compute the cross product of the direction vectors: ${\vec{b}}_{1} \times {\vec{b}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 3 \\ 1 & 4 & - 5 \end{array} ∣ = \hat{i} (- 10 - (- 12)) - \hat{j} (- 5 - (- 3)) + \hat{k} (4 - 2) = 2 \hat{i} + 2 \hat{j} + 2 \hat{k}$ . Magnitude: $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{2^{2} + 2^{2} + 2^{2}} = \sqrt{12} = 2 \sqrt{3}$ .

06Execute

Compute the scalar triple product and the shortest distance: $({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) = (- 8) (2) + (- 7) (2) + (- 3) (2) = - 16 - 14 - 6 = - 36$ . $d = \frac{∣ - 36 ∣}{2 \sqrt{3}} = \frac{36}{2 \sqrt{3}} = \frac{18}{\sqrt{3}} = 6 \sqrt{3}$ .

✓Verify

Check scalar triple product using determinant directly: $∣ \begin{array}{ccc} - 8 & - 7 & - 3 \\ 1 & 2 & - 3 \\ 1 & 4 & - 5 \end{array} ∣ = - 8 (2) - (- 7) (- 2) + (- 3) (2) = - 16 - 14 - 6 = - 36$ . Dividing by $\sqrt{12}$ yields $\frac{36}{\sqrt{12}} = 6 \sqrt{3}$ .

Hints that build this answer step by step

What are the direction vectors ${\vec{b}}_{1}, {\vec{b}}_{2}$ and position vectors ${\vec{a}}_{1}, {\vec{a}}_{2}$ of points lying on the respective lines?
${\vec{a}}_{1} = (5, 2, 4), {\vec{b}}_{1} = (1, 2, - 3)$ and ${\vec{a}}_{2} = (- 3, - 5, 1), {\vec{b}}_{2} = (1, 4, - 5)$
What is the cross product of the direction vectors ${\vec{b}}_{1} \times {\vec{b}}_{2}$ ?
$2 \hat{i} + 2 \hat{j} + 2 \hat{k}$
What is the shortest distance between the lines, calculated using $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ ?
$6 \sqrt{3}$

✓ 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

Why do we take the cross product of the two direction vectors?

The line of shortest distance between two skew lines is perpendicular to both lines, so its direction is along ${\vec{b}}_{1} \times {\vec{b}}_{2}$ .

Answer

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

Why each option works or fails

A: 5√(3) - Evaluating the scalar triple product numerator as $30 \sqrt{3}$ or miscalculating the dot product with the cross product vector leads to an incorrect coefficient of 5. Carefully compute $({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) = - 8 (2) - 7 (2) + 3 (2) = - 36$ , giving absolute value $36$ , which divides by $2 \sqrt{3}$ to give $18 / \sqrt{3} = 6 \sqrt{3}$ .
B: 7√(3) - Making an arithmetic slip when computing the difference vector ${\vec{a}}_{2} - {\vec{a}}_{1}$ or its dot product with the normal vector leads to a numerator of $42$ . Check the coordinates of the reference points: ${\vec{a}}_{1} = (5, 2, 4)$ and ${\vec{a}}_{2} = (- 3, - 5, 1)$ , giving ${\vec{a}}_{2} - {\vec{a}}_{1} = (- 8, - 7, - 3)$ and a numerator magnitude of $36$ , not $42$ .
C · correct: 6√(3) - The student correctly applied the formula $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ . This option is correct.
D: 4√(3) - Miscalculating the cross product magnitude as $\sqrt{27}$ or making an error in simplifying $36 / \sqrt{12}$ leads to $4 \sqrt{3}$ . Evaluate $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{2^{2} + 2^{2} + 2^{2}} = \sqrt{12} = 2 \sqrt{3}$ . Dividing $36$ by $2 \sqrt{3}$ gives $18 / \sqrt{3} = 6 \sqrt{3}$ .

Step-by-step solution

Given: Line $L_{1}$ : point ${\vec{a}}_{1} = (5, 2, 4)$ , direction vector ${\vec{b}}_{1} = \hat{i} + 2 \hat{j} - 3 \hat{k}$ . Line $L_{2}$ : point ${\vec{a}}_{2} = (- 3, - 5, 1)$ , direction vector ${\vec{b}}_{2} = \hat{i} + 4 \hat{j} - 5 \hat{k}$ .
Goal: Find the shortest distance $d = ∣ \frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣} ∣$ between the two skew lines.
Approach: Compute ${\vec{a}}_{2} - {\vec{a}}_{1}$ and ${\vec{b}}_{1} \times {\vec{b}}_{2}$ , then compute their dot product and divide by the magnitude $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣$ .
Execute: Find the vector connecting the points: ${\vec{a}}_{2} - {\vec{a}}_{1} = (- 3 - 5) \hat{i} + (- 5 - 2) \hat{j} + (1 - 4) \hat{k} = - 8 \hat{i} - 7 \hat{j} - 3 \hat{k}$ .
Execute: Compute the cross product of the direction vectors: ${\vec{b}}_{1} \times {\vec{b}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 3 \\ 1 & 4 & - 5 \end{array} ∣ = \hat{i} (- 10 - (- 12)) - \hat{j} (- 5 - (- 3)) + \hat{k} (4 - 2) = 2 \hat{i} + 2 \hat{j} + 2 \hat{k}$ . Magnitude: $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{2^{2} + 2^{2} + 2^{2}} = \sqrt{12} = 2 \sqrt{3}$ .
Execute: Compute the scalar triple product and the shortest distance: $({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) = (- 8) (2) + (- 7) (2) + (- 3) (2) = - 16 - 14 - 6 = - 36$ . $d = \frac{∣ - 36 ∣}{2 \sqrt{3}} = \frac{36}{2 \sqrt{3}} = \frac{18}{\sqrt{3}} = 6 \sqrt{3}$ .
Verify: Check scalar triple product using determinant directly: $∣ \begin{array}{ccc} - 8 & - 7 & - 3 \\ 1 & 2 & - 3 \\ 1 & 4 & - 5 \end{array} ∣ = - 8 (2) - (- 7) (- 2) + (- 3) (2) = - 16 - 14 - 6 = - 36$ . Dividing by $\sqrt{12}$ yields $\frac{36}{\sqrt{12}} = 6 \sqrt{3}$ .

Shortest distance between two skew lines in 3D space

What feels right?

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

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

We think you should solve this next.

A vector $\vec{v}$ in the first octant is inclined to the x-axis at $60^{\circ}$ , to…

Let $(α, β, γ)$ be the image of the point $P (2, 3, 5)$ in the plane…

Another question from the same paper — Integral Calculus, 2023

Students also ask

Shortest distance between two skew lines in 3D space

The shortest distance between the given skew lines is 6363​.

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

A vector v⃗v in the first octant is inclined to the x-axis at 60∘60∘, to…

Let (α,β,γ)(α,β,γ) be the image of the point P(2,3,5)P(2,3,5) in the plane…

Another question from the same paper — Integral Calculus, 2023

Students also ask

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

A vector $\vec{v}$ in the first octant is inclined to the x-axis at $60^{\circ}$ , to…

Let $(α, β, γ)$ be the image of the point $P (2, 3, 5)$ in the plane…