JEE MainMathematics

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Shortest Distance Between Two Skew Lines in 3D Space

If the shortest distance between the line joining the points (1,2,3) and (2,3,4), and the line (x-1)/2=(y+1)/(-1)=(z-2)/0 is a, then 28a^2 is equal to

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Correct answer

The shortest distance squared multiplied by 28 evaluates to 18.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Line $L_{1}$ passes through $A (1, 2, 3)$ and $B (2, 3, 4)$ . Line $L_{2}$ is given by $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ . Shortest distance between $L_{1}$ and $L_{2}$ is $a$ .

02Goal

Find the value of $28 a^{2}$ , where $a$ is the shortest distance between lines $L_{1}$ and $L_{2}$ .

03Approach

Express $L_{1}$ in vector form $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $L_{2}$ as $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ . Then apply the shortest distance formula $a = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ , and compute $28 a^{2}$ .

04Execute

For $L_{1}$ : point ${\vec{a}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and direction vector ${\vec{b}}_{1} = (2 - 1) \hat{i} + (3 - 2) \hat{j} + (4 - 3) \hat{k} = \hat{i} + \hat{j} + \hat{k}$ . For $L_{2}$ : point ${\vec{a}}_{2} = \hat{i} - \hat{j} + 2 \hat{k}$ and direction vector ${\vec{b}}_{2} = 2 \hat{i} - \hat{j} + 0 \hat{k}$ . Compute difference vector: ${\vec{a}}_{2} - {\vec{a}}_{1} = (1 - 1) \hat{i} + (- 1 - 2) \hat{j} + (2 - 3) \hat{k} = - 3 \hat{j} - \hat{k}$ . Compute cross product ${\vec{b}}_{1} \times {\vec{b}}_{2}$ : ${\vec{b}}_{1} \times {\vec{b}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & - 1 & 0 \end{array} ∣ = \hat{i} (0 - (- 1)) - \hat{j} (0 - 2) + \hat{k} (- 1 - 2) = \hat{i} + 2 \hat{j} - 3 \hat{k}$ Magnitude: $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{1^{2} + 2^{2} + (- 3)^{2}} = \sqrt{1 + 4 + 9} = \sqrt{14}$ . Scalar triple product in the numerator: $∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣ = ∣ (- 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 3 \hat{k}) ∣ = ∣ 0 (1) + (- 3) (2) + (- 1) (- 3) ∣ = ∣ - 6 + 3 ∣ = ∣ - 3 ∣ = 3$ Hence, shortest distance $a = \frac{3}{\sqrt{14}}$ . Finally, calculate $28 a^{2} = 28 \times {(\frac{3}{\sqrt{14}})}^{2} = 28 \times \frac{9}{14} = 18$ .

✓Verify

Verify scalar triple product using determinant form directly: $∣ \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ b_{1 x} & b_{1 y} & b_{1 z} \\ b_{2 x} & b_{2 y} & b_{2 z} \end{array} ∣ = ∣ \begin{array}{ccc} 0 & - 3 & - 1 \\ 1 & 1 & 1 \\ 2 & - 1 & 0 \end{array} ∣ = 0 - (- 3) (0 - 2) + (- 1) (- 1 - 2) = - 6 + 3 = - 3$ Absolute value is 3. Denominator $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{14}$ . The result $a^{2} = 9 / 14 ⟹ 28 a^{2} = 18$ is fully confirmed.

Hints that build this answer step by step

What are the direction vector and a point on the line passing through $(1, 2, 3)$ and $(2, 3, 4)$ ?
Passing point ${\vec{a}}_{1} = ⟨ 1, 2, 3 ⟩$ and direction vector ${\vec{b}}_{1} = ⟨ 1, 1, 1 ⟩$
What is the cross product ${\vec{b}}_{1} \times {\vec{b}}_{2}$ of the two direction vectors ${\vec{b}}_{1} = ⟨ 1, 1, 1 ⟩$ and ${\vec{b}}_{2} = ⟨ 2, - 1, 0 ⟩$ ?
$⟨ 1, 2, - 3 ⟩$
Using the line points ${\vec{a}}_{1} = ⟨ 1, 2, 3 ⟩$ and ${\vec{a}}_{2} = ⟨ 1, - 1, 2 ⟩$ , what is the value of the scalar triple product $({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})$ ?
-3
Given the magnitude $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{1^{2} + 2^{2} + (- 3)^{2}} = \sqrt{14}$ , what is the value of $28 a^{2}$ ?
18

✓ 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: 19 September 2026

Quick checks

Students also ask

Why does the z-component of the direction vector of L2 have 0 in the denominator?

A zero in the denominator means the direction component along that axis is zero. This means the line lies in a plane where $z$ is constant. Here, $z - 2 = 0$ , so $z = 2$ .

Answer

Accepted answer: 18

The shortest distance squared multiplied by 28 evaluates to 18.

Why each option works or fails

Step-by-step solution

Given: Line $L_{1}$ passes through $A (1, 2, 3)$ and $B (2, 3, 4)$ . Line $L_{2}$ is given by $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ . Shortest distance between $L_{1}$ and $L_{2}$ is $a$ .
Goal: Find the value of $28 a^{2}$ , where $a$ is the shortest distance between lines $L_{1}$ and $L_{2}$ .
Approach: Express $L_{1}$ in vector form $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $L_{2}$ as $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ . Then apply the shortest distance formula $a = \frac{∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ , and compute $28 a^{2}$ .
Execute: For $L_{1}$ : point ${\vec{a}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and direction vector ${\vec{b}}_{1} = (2 - 1) \hat{i} + (3 - 2) \hat{j} + (4 - 3) \hat{k} = \hat{i} + \hat{j} + \hat{k}$ . For $L_{2}$ : point ${\vec{a}}_{2} = \hat{i} - \hat{j} + 2 \hat{k}$ and direction vector ${\vec{b}}_{2} = 2 \hat{i} - \hat{j} + 0 \hat{k}$ . Compute difference vector: ${\vec{a}}_{2} - {\vec{a}}_{1} = (1 - 1) \hat{i} + (- 1 - 2) \hat{j} + (2 - 3) \hat{k} = - 3 \hat{j} - \hat{k}$ . Compute cross product ${\vec{b}}_{1} \times {\vec{b}}_{2}$ : ${\vec{b}}_{1} \times {\vec{b}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & - 1 & 0 \end{array} ∣ = \hat{i} (0 - (- 1)) - \hat{j} (0 - 2) + \hat{k} (- 1 - 2) = \hat{i} + 2 \hat{j} - 3 \hat{k}$ Magnitude: $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{1^{2} + 2^{2} + (- 3)^{2}} = \sqrt{1 + 4 + 9} = \sqrt{14}$ . Scalar triple product in the numerator: $∣ ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) ∣ = ∣ (- 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 3 \hat{k}) ∣ = ∣ 0 (1) + (- 3) (2) + (- 1) (- 3) ∣ = ∣ - 6 + 3 ∣ = ∣ - 3 ∣ = 3$ Hence, shortest distance $a = \frac{3}{\sqrt{14}}$ . Finally, calculate $28 a^{2} = 28 \times {(\frac{3}{\sqrt{14}})}^{2} = 28 \times \frac{9}{14} = 18$ .
Verify: Verify scalar triple product using determinant form directly: $∣ \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ b_{1 x} & b_{1 y} & b_{1 z} \\ b_{2 x} & b_{2 y} & b_{2 z} \end{array} ∣ = ∣ \begin{array}{ccc} 0 & - 3 & - 1 \\ 1 & 1 & 1 \\ 2 & - 1 & 0 \end{array} ∣ = 0 - (- 3) (0 - 2) + (- 1) (- 1 - 2) = - 6 + 3 = - 3$ Absolute value is 3. Denominator $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣ = \sqrt{14}$ . The result $a^{2} = 9 / 14 ⟹ 28 a^{2} = 18$ is fully confirmed.

Shortest Distance Between Two Skew Lines in 3D Space

What feels right?

The shortest distance squared multiplied by 28 evaluates to 18.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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