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Three Dimensional Geometry: Mathematics | JEE Main

The shortest distance between the lines

\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-6}{2}

and

\frac{x-6}{3} = \frac{1-y}{2} = \frac{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 $\vec{b} = \langle 6, 1, -8 \rangle$ with direction vector $\vec{q} = \langle 3, -2, 0 \rangle$ . Line $L_1$ passes through $\vec{a} = \langle 2, -1, 6 \rangle$ with direction vector $\vec{p} = \langle 3, 2, 2 \rangle$ . The shortest distance is $d = \frac{|(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$ .

03execute

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

04execute

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

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 $\vec{n} = \langle 2, 3, -6 \rangle$ : $|\vec{n}| = 7$ , 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 $\vec{r} = \vec{a}_i + \lambda \vec{b}_i$ of the two lines?
Line 1 passes through $(2, -1, 6)$ with direction $\langle 3, 2, 2 \rangle$ ; Line 2 passes through $(6, 1, -8)$ with direction $\langle 3, -2, 0 \rangle$ .
What is the cross product of the direction vectors $\vec{b}_1 = \langle 3, 2, 2 \rangle$ and $\vec{b}_2 = \langle 3, -2, 0 \rangle$ ?
$\langle 4, 6, -12 \rangle$
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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $\vec{b} = \langle 6, 1, -8 \rangle$ with direction vector $\vec{q} = \langle 3, -2, 0 \rangle$ . Line $L_1$ passes through $\vec{a} = \langle 2, -1, 6 \rangle$ with direction vector $\vec{p} = \langle 3, 2, 2 \rangle$ . The shortest distance is $d = \frac{|(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$ .
execute: Compute $\vec{b} - \vec{a} = \langle 6-2, 1 - (-1), -8 - 6 \rangle = \langle 4, 2, -14 \rangle$ .
execute: Compute $\vec{p} \times \vec{q} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{matrix} \right| = \hat{i}(0 - (-4)) - \hat{j}(0 - 6) + \hat{k}(-6 - 6) = 4\hat{i} + 6\hat{j} - 12\hat{k}$ .
execute: Evaluate the scalar triple product: $(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) = 4(4) + 2(6) + (-14)(-12) = 16 + 12 + 168 = 196$ . 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 $\vec{n} = \langle 2, 3, -6 \rangle$ : $|\vec{n}| = 7$ , 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.

Three Dimensional Geometry: Mathematics | JEE Main

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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