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Three Dimensional Geometry: Mathematics | JEE Main

Q: Can we rescale direction vectors by multiplying by integers?

Yes, because the cross product and its norm rescale by the same positive constant factor, which cancels out completely in the formula for .

The shortest distance between the lines

x + 1 = 2y = -12z

and

x = y + 2 = 6z - 6

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What feels right?

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\frac{3}{2}

B2C

\frac{5}{2}

Step-by-step solutionView

Correct answer

The shortest distance between the two lines is 2.

Option analysis

Why each option works or fails

A ·

\frac{3}{2}

Computing the numerator $|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|$ or the magnitude $|\mathbf{b}_1 \times \mathbf{b}_2|$ with an arithmetic error, leading to an incorrect ratio of $\frac{3}{2}$ . Double-check vector cross products and scalar dot products carefully: the scalar triple product evaluates to 12 and the magnitude of the normal vector is 6, yielding $12 / 6 = 2$ .

B · 2

Correct option. Writing both lines in standard symmetric form gives points $\mathbf{a}_1 = (-1, 0, 0)$ , $\mathbf{a}_2 = (0, -2, 1)$ and direction vectors proportional to $\mathbf{b}_1 = \langle 6, 3, -1 \rangle$ , $\mathbf{b}_2 = \langle 6, 6, 1 \rangle$ . The shortest distance formula $d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$ gives $\frac{|9 - 6 + 9|}{\sqrt{9 + 144 + 324}/3} = \frac{12}{6} = 2$ .

C ·

\frac{5}{2}

Incorrectly rewriting the standard symmetric form of $6z - 6$ as $6(z - 1)$ resulting in a point coordinate error, such as taking $z_2 = -1$ instead of $1$ . Ensure expressions like $6z - 6$ are factored as $6(z - 1)$ so the numerator becomes $z - 1$ , giving point coordinate $z = 1$ .

D · 3

Misidentifying the direction ratios by inverting coefficients incorrectly, such as taking coefficients directly as direction ratios instead of their reciprocals. To find direction ratios from $x + 1 = 2y = -12z$ , divide by a common multiple like 12 to get $\frac{x+1}{12} = \frac{y}{6} = \frac{z}{-1}$ , giving direction vector $\langle 12, 6, -1 \rangle$ .

Reviewed route

Solution

StepWorking

01given

Two lines: $L_1: x + 1 = 2y = -12z$ and $L_2: x = y + 2 = 6z - 6$ .

02goal

Find the shortest distance $d$ between lines $L_1$ and $L_2$ .

03approach

Rewrite both lines in standard symmetric Cartesian form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ to identify base points $\vec{a}, \vec{b}$ and direction vectors $\vec{p}, \vec{q}$ . Clear fractions in direction vectors for simpler cross products, then apply $d = \frac{|(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$ .

04execute

Express $L_1$ in standard form: $\frac{x - (-1)}{1} = \frac{y - 0}{1/2} = \frac{z - 0}{-1/12}$ , so point $A = (-1, 0, 0)$ and direction vector $\vec{p} = 12\left(1, \frac{1}{2}, -\frac{1}{12}\right) = \langle 12, 6, -1 \rangle$ . Express $L_2$ in standard form: $\frac{x - 0}{1} = \frac{y - (-2)}{1} = \frac{z - 1}{1/6}$ , so point $B = (0, -2, 1)$ and direction vector $\vec{q} = 6\left(1, 1, \frac{1}{6}\right) = \langle 6, 6, 1 \rangle$ .

05execute

Find $\vec{b} - \vec{a} = \langle 0 - (-1), -2 - 0, 1 - 0 \rangle = \langle 1, -2, 1 \rangle$ . Compute the normal vector $\vec{n} = \vec{p} \times \vec{q}$ : $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} = \hat{i}(6 - (-6)) - \hat{j}(12 - (-6)) + \hat{k}(72 - 36) = 12\hat{i} - 18\hat{j} + 36\hat{k} = 6\langle 2, -3, 6 \rangle$ .

06execute

Calculate the magnitude of $\langle 2, -3, 6 \rangle$ : $|\langle 2, -3, 6 \rangle| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ . Compute the scalar triple product component: $(\vec{b} - \vec{a}) \cdot \langle 2, -3, 6 \rangle = 1(2) + (-2)(-3) + 1(6) = 2 + 6 + 6 = 14$ . Then $d = \frac{|14|}{7} = 2$ .

✓verify

Check scalar triple product directly via determinant: $\begin{vmatrix} 1 & -2 & 1 \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} = 1(12) - (-2)(18) + 1(36) = 12 + 36 + 36 = 84$ . $|\vec{p} \times \vec{q}| = 6 \times 7 = 42$ . $d = \frac{84}{42} = 2$ . Consistent and positive.

Hints that build this answer step by step

What are the standard symmetric forms, a point, and a direction vector for each of the two lines?
Line 1: passes through $(-1, 0, 0)$ with direction $\langle 6, 3, -0.5 \rangle$ ; Line 2: passes through $(0, -2, 1)$ with direction $\langle 6, 6, 1 \rangle$
Using direction vectors $\mathbf{b}_1 = \langle 12, 6, -1 \rangle$ and $\mathbf{b}_2 = \langle 6, 6, 1 \rangle$ , what is the cross product $\mathbf{b}_1 \times \mathbf{b}_2$ and its magnitude?
$\mathbf{b}_1 \times \mathbf{b}_2 = \langle 12, -18, 36 \rangle$ with magnitude $\sqrt{12^2 + (-18)^2 + 36^2} = 42$
With $\mathbf{a}_2 - \mathbf{a}_1 = \langle 0 - (-1), -2 - 0, 1 - 0 \rangle = \langle 1, -2, 1 \rangle$ , what is the shortest distance $d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$ ?
$d = \frac{|1(12) + (-2)(-18) + 1(36)|}{42} = \frac{84}{42} = 2$

✓ 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: 9 September 2026

Quick checks

Students also ask

Can we rescale direction vectors by multiplying by integers?

Yes, because the cross product $\vec{p} \times \vec{q}$ and its norm rescale by the same positive constant factor, which cancels out completely in the formula for $d$ .

Answer

The shortest distance between the two lines is 2.

Why each option works or fails

A: $\frac{3}{2}$ - Computing the numerator $|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|$ or the magnitude $|\mathbf{b}_1 \times \mathbf{b}_2|$ with an arithmetic error, leading to an incorrect ratio of $\frac{3}{2}$ . Double-check vector cross products and scalar dot products carefully: the scalar triple product evaluates to 12 and the magnitude of the normal vector is 6, yielding $12 / 6 = 2$ .
B · correct: 2 - Correct option. Writing both lines in standard symmetric form gives points $\mathbf{a}_1 = (-1, 0, 0)$ , $\mathbf{a}_2 = (0, -2, 1)$ and direction vectors proportional to $\mathbf{b}_1 = \langle 6, 3, -1 \rangle$ , $\mathbf{b}_2 = \langle 6, 6, 1 \rangle$ . The shortest distance formula $d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$ gives $\frac{|9 - 6 + 9|}{\sqrt{9 + 144 + 324}/3} = \frac{12}{6} = 2$ .
C: $\frac{5}{2}$ - Incorrectly rewriting the standard symmetric form of $6z - 6$ as $6(z - 1)$ resulting in a point coordinate error, such as taking $z_2 = -1$ instead of $1$ . Ensure expressions like $6z - 6$ are factored as $6(z - 1)$ so the numerator becomes $z - 1$ , giving point coordinate $z = 1$ .
D: 3 - Misidentifying the direction ratios by inverting coefficients incorrectly, such as taking coefficients directly as direction ratios instead of their reciprocals. To find direction ratios from $x + 1 = 2y = -12z$ , divide by a common multiple like 12 to get $\frac{x+1}{12} = \frac{y}{6} = \frac{z}{-1}$ , giving direction vector $\langle 12, 6, -1 \rangle$ .

Step-by-step solution

given: Two lines: $L_1: x + 1 = 2y = -12z$ and $L_2: x = y + 2 = 6z - 6$ .
goal: Find the shortest distance $d$ between lines $L_1$ and $L_2$ .
approach: Rewrite both lines in standard symmetric Cartesian form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ to identify base points $\vec{a}, \vec{b}$ and direction vectors $\vec{p}, \vec{q}$ . Clear fractions in direction vectors for simpler cross products, then apply $d = \frac{|(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$ .
execute: Express $L_1$ in standard form: $\frac{x - (-1)}{1} = \frac{y - 0}{1/2} = \frac{z - 0}{-1/12}$ , so point $A = (-1, 0, 0)$ and direction vector $\vec{p} = 12\left(1, \frac{1}{2}, -\frac{1}{12}\right) = \langle 12, 6, -1 \rangle$ . Express $L_2$ in standard form: $\frac{x - 0}{1} = \frac{y - (-2)}{1} = \frac{z - 1}{1/6}$ , so point $B = (0, -2, 1)$ and direction vector $\vec{q} = 6\left(1, 1, \frac{1}{6}\right) = \langle 6, 6, 1 \rangle$ .
execute: Find $\vec{b} - \vec{a} = \langle 0 - (-1), -2 - 0, 1 - 0 \rangle = \langle 1, -2, 1 \rangle$ . Compute the normal vector $\vec{n} = \vec{p} \times \vec{q}$ : $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} = \hat{i}(6 - (-6)) - \hat{j}(12 - (-6)) + \hat{k}(72 - 36) = 12\hat{i} - 18\hat{j} + 36\hat{k} = 6\langle 2, -3, 6 \rangle$ .
execute: Calculate the magnitude of $\langle 2, -3, 6 \rangle$ : $|\langle 2, -3, 6 \rangle| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ . Compute the scalar triple product component: $(\vec{b} - \vec{a}) \cdot \langle 2, -3, 6 \rangle = 1(2) + (-2)(-3) + 1(6) = 2 + 6 + 6 = 14$ . Then $d = \frac{|14|}{7} = 2$ .
verify: Check scalar triple product directly via determinant: $\begin{vmatrix} 1 & -2 & 1 \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} = 1(12) - (-2)(18) + 1(36) = 12 + 36 + 36 = 84$ . $|\vec{p} \times \vec{q}| = 6 \times 7 = 42$ . $d = \frac{84}{42} = 2$ . Consistent and positive.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The shortest distance between the two lines is 2.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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