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Three Dimensional Geometry: Mathematics | JEE Main

The shortest distance between the lines

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

and

\frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}

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5\sqrt{3}

2\sqrt{3}

3\sqrt{3}

4\sqrt{3}

Step-by-step solutionView

Correct answer

The shortest distance between the given skew lines is $4\sqrt{3}$ .

Option analysis

Why each option works or fails

A ·

5\sqrt{3}

Computing the scalar triple product $\lvert(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)\rvert$ as $80$ instead of $64$ , or making an arithmetic error when evaluating coordinates. Check the evaluation of the scalar triple product carefully: $(0)(-16) - (-10)(-16) + (-2)(-16) = -160 + 32 = -128$ , giving an absolute value of $128$ .

B ·

2\sqrt{3}

Misidentifying the point on the second line as $(1, 8, 4)$ instead of $(1, -8, 4)$ due to a sign error in the term $(y+8)$ . Write the symmetric form as $\frac{y - y_1}{b_y}$ ; thus $\frac{y+8}{-7}$ implies $y_1 = -8$ , not $+8$ .

C ·

3\sqrt{3}

Arithmetic slip when simplifying the ratio $\frac{128}{16\sqrt{3}} = \frac{8}{\sqrt{3}}$ , leading to miscalculating the numerator or denominator factors. Simplify $\frac{128}{16\sqrt{3}}$ directly: $128 / 16 = 8$ , so the distance is $\frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3}$ , or check for arithmetic slips in computing $\lvert \vec{b}_1 \times \vec{b}_2 \rvert$ .

D ·

4\sqrt{3}

None. The cross product of the direction vectors is $(-16, -16, -16)$ with magnitude $16\sqrt{3}$ , and the scalar triple product with $\vec{a}_2 - \vec{a}_1 = (0, -10, -2)$ gives $128$ , yielding $\frac{128}{16\sqrt{3}} = \frac{8}{\sqrt{3}}$ (or with correct cross product $(-16, -16, -16)$ , the distance is $4\sqrt{3}$ when evaluating $(0, -10, -2) \cdot (-16, -16, -16)$ ). Correctly apply the formula $d = \frac{\lvert(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)\rvert}{\lvert\vec{b}_1 \times \vec{b}_2\rvert}$ .

Reviewed route

Solution

StepWorking

01given

Line 1: passes through $\vec{a}_1 = (1, 2, 6)$ with direction $\vec{b}_1 = \langle 2, 1, -3 \rangle$ . Line 2: passes through $\vec{a}_2 = (1, -8, 4)$ with direction $\vec{b}_2 = \langle 2, -7, 5 \rangle$ .

02goal

Find the shortest distance $d$ between the two skew lines.

03approach

Use the standard skew-distance formula: $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ . Compute the cross product of the direction vectors, find its magnitude, and evaluate the projection of the displacement vector along the common normal.

04execute

Compute $\vec{b}_1 \times \vec{b}_2$ : $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix} = \hat{i}(5 - 21) - \hat{j}(10 - (-6)) + \hat{k}(-14 - 2) = -16\hat{i} - 16\hat{j} - 16\hat{k} = -16(\hat{i} + \hat{j} + \hat{k})$ Its magnitude is $|\vec{b}_1 \times \vec{b}_2| = 16\sqrt{1^2 + 1^2 + 1^2} = 16\sqrt{3}$ . Now compute $\vec{a}_2 - \vec{a}_1 = (1 - 1)\hat{i} + (-8 - 2)\hat{j} + (4 - 6)\hat{k} = -10\hat{j} - 2\hat{k}$ . The numerator scalar triple product is: $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)| = |(-10\hat{j} - 2\hat{k}) \cdot (-16)(\hat{i} + \hat{j} + \hat{k})| = 16 |0 - 10 - 2| = 16 \times 12 = 192$ Thus, $d = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$ .

✓verify

$12 / \sqrt{3} = 12\sqrt{3}/3 = 4\sqrt{3}$ . Since $\vec{n} = \frac{1}{\sqrt{3}}(1, 1, 1)$ , the projection of $\vec{a}_1 - \vec{a}_2 = (0, 10, 2)$ onto $\vec{n}$ is $(0 + 10 + 2)/\sqrt{3} = 12/\sqrt{3} = 4\sqrt{3}$ , matching option (3).

Hints that build this answer step by step

What are the position vectors $\vec{a}_1, \vec{a}_2$ of points on each line, and their direction vectors $\vec{b}_1, \vec{b}_2$ ?
$\vec{a}_1 = (1, 2, 6)$ , $\vec{b}_1 = (2, 1, -3)$ and $\vec{a}_2 = (1, -8, 4)$ , $\vec{b}_2 = (2, -7, 5)$
What is the cross product of the direction vectors $\vec{b}_1 \times \vec{b}_2$ ?
$-16\hat{i} - 16\hat{j} - 16\hat{k}$
What is the shortest distance between the two lines using $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ ?
$4\sqrt{3}$

✓ 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 the sign of the cross product matter for shortest distance?

No. The formula uses the absolute value of the scalar triple product in the numerator. The denominator contains the magnitude of the normal vector. Therefore, distance is always non-negative.

Answer

The shortest distance between the given skew lines is $4\sqrt{3}$ .

Why each option works or fails

A: $5\sqrt{3}$ - Computing the scalar triple product $\lvert(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)\rvert$ as $80$ instead of $64$ , or making an arithmetic error when evaluating coordinates. Check the evaluation of the scalar triple product carefully: $(0)(-16) - (-10)(-16) + (-2)(-16) = -160 + 32 = -128$ , giving an absolute value of $128$ .
B: $2\sqrt{3}$ - Misidentifying the point on the second line as $(1, 8, 4)$ instead of $(1, -8, 4)$ due to a sign error in the term $(y+8)$ . Write the symmetric form as $\frac{y - y_1}{b_y}$ ; thus $\frac{y+8}{-7}$ implies $y_1 = -8$ , not $+8$ .
C: $3\sqrt{3}$ - Arithmetic slip when simplifying the ratio $\frac{128}{16\sqrt{3}} = \frac{8}{\sqrt{3}}$ , leading to miscalculating the numerator or denominator factors. Simplify $\frac{128}{16\sqrt{3}}$ directly: $128 / 16 = 8$ , so the distance is $\frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3}$ , or check for arithmetic slips in computing $\lvert \vec{b}_1 \times \vec{b}_2 \rvert$ .
D · correct: $4\sqrt{3}$ - None. The cross product of the direction vectors is $(-16, -16, -16)$ with magnitude $16\sqrt{3}$ , and the scalar triple product with $\vec{a}_2 - \vec{a}_1 = (0, -10, -2)$ gives $128$ , yielding $\frac{128}{16\sqrt{3}} = \frac{8}{\sqrt{3}}$ (or with correct cross product $(-16, -16, -16)$ , the distance is $4\sqrt{3}$ when evaluating $(0, -10, -2) \cdot (-16, -16, -16)$ ). Correctly apply the formula $d = \frac{\lvert(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)\rvert}{\lvert\vec{b}_1 \times \vec{b}_2\rvert}$ .

Step-by-step solution

given: Line 1: passes through $\vec{a}_1 = (1, 2, 6)$ with direction $\vec{b}_1 = \langle 2, 1, -3 \rangle$ . Line 2: passes through $\vec{a}_2 = (1, -8, 4)$ with direction $\vec{b}_2 = \langle 2, -7, 5 \rangle$ .
goal: Find the shortest distance $d$ between the two skew lines.
approach: Use the standard skew-distance formula: $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ . Compute the cross product of the direction vectors, find its magnitude, and evaluate the projection of the displacement vector along the common normal.
execute: Compute $\vec{b}_1 \times \vec{b}_2$ : $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix} = \hat{i}(5 - 21) - \hat{j}(10 - (-6)) + \hat{k}(-14 - 2) = -16\hat{i} - 16\hat{j} - 16\hat{k} = -16(\hat{i} + \hat{j} + \hat{k})$ Its magnitude is $|\vec{b}_1 \times \vec{b}_2| = 16\sqrt{1^2 + 1^2 + 1^2} = 16\sqrt{3}$ . Now compute $\vec{a}_2 - \vec{a}_1 = (1 - 1)\hat{i} + (-8 - 2)\hat{j} + (4 - 6)\hat{k} = -10\hat{j} - 2\hat{k}$ . The numerator scalar triple product is: $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)| = |(-10\hat{j} - 2\hat{k}) \cdot (-16)(\hat{i} + \hat{j} + \hat{k})| = 16 |0 - 10 - 2| = 16 \times 12 = 192$ Thus, $d = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$ .
verify: $12 / \sqrt{3} = 12\sqrt{3}/3 = 4\sqrt{3}$ . Since $\vec{n} = \frac{1}{\sqrt{3}}(1, 1, 1)$ , the projection of $\vec{a}_1 - \vec{a}_2 = (0, 10, 2)$ onto $\vec{n}$ is $(0 + 10 + 2)/\sqrt{3} = 12/\sqrt{3} = 4\sqrt{3}$ , matching option (3).

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The shortest distance between the given skew lines is $4\sqrt{3}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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