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Three Dimensional Geometry: Mathematics | JEE Main

The shortest distance between the lines

\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}

and

\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}

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What feels right?

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3 \sqrt{6}

6 \sqrt{3}

6 \sqrt{2}

2 \sqrt{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 \sqrt{6}

None. 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 \sqrt{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 \sqrt{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 \sqrt{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: $\vec{a}_1 = (4, -2, -3)$ , $\vec{b}_1 = \langle 4, 5, 3 \rangle$ ; Line 2: $\vec{a}_2 = (1, 3, 4)$ , $\vec{b}_2 = \langle 3, 4, 2 \rangle$ .

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 $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix} = \hat{i}(10 - 12) - \hat{j}(8 - 9) + \hat{k}(16 - 15) = -2\hat{i} + \hat{j} + \hat{k}$ . 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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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: $3 \sqrt{6}$ - None. 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: $6 \sqrt{3}$ - 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: $6 \sqrt{2}$ - 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 $6 \sqrt{3}$ 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: $2 \sqrt{6}$ - 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: $\vec{a}_1 = (4, -2, -3)$ , $\vec{b}_1 = \langle 4, 5, 3 \rangle$ ; Line 2: $\vec{a}_2 = (1, 3, 4)$ , $\vec{b}_2 = \langle 3, 4, 2 \rangle$ .
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 $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix} = \hat{i}(10 - 12) - \hat{j}(8 - 9) + \hat{k}(16 - 15) = -2\hat{i} + \hat{j} + \hat{k}$ . 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.

Three Dimensional Geometry: Mathematics | JEE Main

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