JEE MainMathematics

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Shortest distance between two skew lines in 3D

The shortest distance between the lines (x-2)/3 = (y+1)/2 = (z-6)/2 and (x-6)/3 = (1-y)/2 = (z+8)/0 is equal to

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Correct answer

The shortest distance between the given lines is 14.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Line $L_{1} : \frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and Line $L_{2} : \frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ .

02Approach

Rewrite the second line in standard symmetric form: $\frac{1 - y}{2} = \frac{y - 1}{- 2}$ , so $L_{2}$ passes through $\vec{b} = ⟨ 6, 1, - 8 ⟩$ with direction vector $\vec{q} = ⟨ 3, - 2, 0 ⟩$ . Line $L_{1}$ passes through $\vec{a} = ⟨ 2, - 1, 6 ⟩$ with direction vector $\vec{p} = ⟨ 3, 2, 2 ⟩$ . The shortest distance is $d = \frac{∣ (\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) ∣}{∣ \vec{p} \times \vec{q} ∣}$ .

03Execute

Compute $\vec{b} - \vec{a} = ⟨ 6 - 2, 1 - (- 1), - 8 - 6 ⟩ = ⟨ 4, 2, - 14 ⟩$ .

04Execute

Compute $\vec{p} \times \vec{q} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{array} ∣ = \hat{i} (0 - (- 4)) - \hat{j} (0 - 6) + \hat{k} (- 6 - 6) = 4 \hat{i} + 6 \hat{j} - 12 \hat{k}$ .

05Execute

Evaluate the scalar triple product: $(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) = 4 (4) + 2 (6) + (- 14) (- 12) = 16 + 12 + 168 = 196$ . The magnitude is $∣ \vec{p} \times \vec{q} ∣ = \sqrt{4^{2} + 6^{2} + (- 12)^{2}} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$ . Thus, $d = \frac{196}{14} = 14$ .

✓Verify

Using simplified direction vector of cross product $\vec{n} = ⟨ 2, 3, - 6 ⟩$ : $∣ \vec{n} ∣ = 7$ , and $(\vec{b} - \vec{a}) \cdot \vec{n} = 4 (2) + 2 (3) + (- 14) (- 6) = 8 + 6 + 84 = 98$ . Then $d = \frac{98}{7} = 14$ . Both lead to 14.

Hints that build this answer step by step

What are the standard vector forms $\vec{r} = {\vec{a}}_{i} + λ {\vec{b}}_{i}$ of the two lines?
Line 1 passes through $(2, - 1, 6)$ with direction $⟨ 3, 2, 2 ⟩$ ; Line 2 passes through $(6, 1, - 8)$ with direction $⟨ 3, - 2, 0 ⟩$ .
What is the cross product of the direction vectors ${\vec{b}}_{1} = ⟨ 3, 2, 2 ⟩$ and ${\vec{b}}_{2} = ⟨ 3, - 2, 0 ⟩$ ?
$⟨ 4, 6, - 12 ⟩$
Using the shortest distance formula $d = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ , what is the final value of $d$ ?
14

✓ Source and academic review

Question type: Numerical
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

Why did the sign of the y-component flip in line 2?

In the standard symmetric form, the coefficient of each coordinate variable (x, y, z) in the numerator must be +1. Since the numerator has (1 - y), we rewrite it as -(y - 1), transferring the negative sign to the denominator: (y - 1)/(-2).

Answer

The shortest distance between the given lines is 14.

Why each option works or fails

Step-by-step solution

Given: Line $L_{1} : \frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and Line $L_{2} : \frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ .
Approach: Rewrite the second line in standard symmetric form: $\frac{1 - y}{2} = \frac{y - 1}{- 2}$ , so $L_{2}$ passes through $\vec{b} = ⟨ 6, 1, - 8 ⟩$ with direction vector $\vec{q} = ⟨ 3, - 2, 0 ⟩$ . Line $L_{1}$ passes through $\vec{a} = ⟨ 2, - 1, 6 ⟩$ with direction vector $\vec{p} = ⟨ 3, 2, 2 ⟩$ . The shortest distance is $d = \frac{∣ (\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) ∣}{∣ \vec{p} \times \vec{q} ∣}$ .
Execute: Compute $\vec{b} - \vec{a} = ⟨ 6 - 2, 1 - (- 1), - 8 - 6 ⟩ = ⟨ 4, 2, - 14 ⟩$ .
Execute: Compute $\vec{p} \times \vec{q} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{array} ∣ = \hat{i} (0 - (- 4)) - \hat{j} (0 - 6) + \hat{k} (- 6 - 6) = 4 \hat{i} + 6 \hat{j} - 12 \hat{k}$ .
Execute: Evaluate the scalar triple product: $\vec{a} = ⟨ 2, - 1, 6 ⟩$ . The magnitude is $∣ \vec{p} \times \vec{q} ∣ = \sqrt{4^{2} + 6^{2} + (- 12)^{2}} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$ . Thus, $d = \frac{196}{14} = 14$ .
Verify: Using simplified direction vector of cross product $\vec{n} = ⟨ 2, 3, - 6 ⟩$ : $∣ \vec{n} ∣ = 7$ , and $(\vec{b} - \vec{a}) \cdot \vec{n} = 4 (2) + 2 (3) + (- 14) (- 6) = 8 + 6 + 84 = 98$ . Then $d = \frac{98}{7} = 14$ . Both lead to 14.

Shortest distance between two skew lines in 3D

What feels right?

The shortest distance between the given lines is 14.

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

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