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Three Dimensional Geometry: Mathematics | JEE Main

Q: Why are the direction ratios of the line the coefficients of the plane equation?

A line perpendicular to a plane has its direction parallel to the normal vector of the plane, which is .

The line

l_1

passes through the point

(2,6,2)

and is perpendicular to the plane

2x + y - 2z = 10

. Then the shortest distance between the line

l_1

and the line

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

is :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{13}{3}

\frac{19}{3}

C7D9

Step-by-step solutionView

Correct answer

The shortest distance between the line $l_1$ and the given line is 9.

Option analysis

Why each option works or fails

A ·

\frac{13}{3}

Computing the numerator using an incorrect dot product or miscalculating the cross product magnitude leads to 13/3. Ensure the cross product of the direction vectors is $\mathbf{b}_1 \times \mathbf{b}_2 = (-4, -8, -8)$ with magnitude $\sqrt{16+64+64}=12$ , and the vector connecting points is $(-3, -10, -2)$ .

B ·

\frac{19}{3}

Dividing the scalar triple product by the wrong vector magnitude or making an arithmetic error when evaluating $(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)$ . Check each component of the dot product: $(-3)(-4) + (-10)(-8) + (-2)(-8) = 12 + 80 + 16 = 108$ .

C · 7

Evaluating the scalar triple product as 84 instead of 108 before dividing by 12. Carefully compute $(-3)(-4) + (-10)(-8) + (-2)(-8) = 108$ , then divide by 12 to get $108 / 12 = 9$ .

D · 9

Correct option: The vector between points is $\mathbf{a}_2 - \mathbf{a}_1 = (-3, -10, -2)$ , and the cross product of direction vectors is $(-4, -8, -8)$ with magnitude 12, yielding a distance of $108/12 = 9$ . None. The shortest distance formula $d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$ was evaluated correctly.

Reviewed route

Solution

StepWorking

01given

Line $l_1$ passes through $A(2, 6, 2)$ and is normal to $2x + y - 2z = 10$ . Line $l_2$ is given by $\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$ .

02goal

Find the shortest distance between skew lines $l_1$ and $l_2$ .

03approach

The direction ratios of $l_1$ are along the normal to the plane: $\vec{p} = \langle 2, 1, -2 \rangle$ . Line $l_2$ passes through $B(-1, -4, 0)$ with direction $\vec{q} = \langle 2, -3, 2 \rangle$ . Shortest distance is $d = \frac{|(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$ .

04execute

Compute $\vec{p} \times \vec{q}$ : $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(4 - (-4)) + \hat{k}(-6 - 2) = -4\hat{i} - 8\hat{j} - 8\hat{k}$ . Its magnitude is $|\vec{p} \times \vec{q}| = \sqrt{(-4)^2 + (-8)^2 + (-8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$ . Compute $\vec{a} - \vec{b} = \langle 2 - (-1), 6 - (-4), 2 - 0 \rangle = \langle 3, 10, 2 \rangle$ . Compute the scalar triple product: $(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q}) = 3(-4) + 10(-8) + 2(-8) = -12 - 80 - 16 = -108$ . Therefore, $d = \frac{|-108|}{12} = 9$ .

✓verify

Scalar determinant method directly: $\begin{vmatrix} 3 & 10 & 2 \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix} = 3(2 - 6) - 10(4 - (-4)) + 2(-6 - 2) = 3(-4) - 10(8) + 2(-8) = -12 - 80 - 16 = -108$ . Denominator is $|\langle -4, -8, -8 \rangle| = 12$ . Ratio is $108 / 12 = 9$ . Correct.

Hints that build this answer step by step

What are the direction vector $\mathbf{b}_1$ and point $\mathbf{a}_1$ of line $l_1$ , given that it passes through $(2,6,2)$ and is perpendicular to the plane $2x + y - 2z = 10$ ?
$\mathbf{a}_1 = (2, 6, 2)$ and $\mathbf{b}_1 = (2, 1, -2)$
What is the cross product $\mathbf{b}_1 \times \mathbf{b}_2$ of the direction vectors $\mathbf{b}_1 = (2, 1, -2)$ and $\mathbf{b}_2 = (2, -3, 2)$ ?
$(-4, -8, -8)$
Using $\mathbf{a}_1 = (2, 6, 2)$ , $\mathbf{a}_2 = (-1, -4, 0)$ , and $\mathbf{b}_1 \times \mathbf{b}_2 = (-4, -8, -8)$ , what is the shortest distance $d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$ ?
9

✓ 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

Why are the direction ratios of the line the coefficients of the plane equation?

A line perpendicular to a plane $Ax + By + Cz = D$ has its direction parallel to the normal vector of the plane, which is $\langle A, B, C \rangle$ .

Answer

The shortest distance between the line $l_1$ and the given line is 9.

Why each option works or fails

A: $\frac{13}{3}$ - Computing the numerator using an incorrect dot product or miscalculating the cross product magnitude leads to 13/3. Ensure the cross product of the direction vectors is $\mathbf{b}_1 \times \mathbf{b}_2 = (-4, -8, -8)$ with magnitude $\sqrt{16+64+64}=12$ , and the vector connecting points is $(-3, -10, -2)$ .
B: $\frac{19}{3}$ - Dividing the scalar triple product by the wrong vector magnitude or making an arithmetic error when evaluating $(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)$ . Check each component of the dot product: $(-3)(-4) + (-10)(-8) + (-2)(-8) = 12 + 80 + 16 = 108$ .
C: 7 - Evaluating the scalar triple product as 84 instead of 108 before dividing by 12. Carefully compute $(-3)(-4) + (-10)(-8) + (-2)(-8) = 108$ , then divide by 12 to get $108 / 12 = 9$ .
D · correct: 9 - Correct option: The vector between points is $\mathbf{a}_2 - \mathbf{a}_1 = (-3, -10, -2)$ , and the cross product of direction vectors is $(-4, -8, -8)$ with magnitude 12, yielding a distance of $108/12 = 9$ . None. The shortest distance formula $d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$ was evaluated correctly.

Step-by-step solution

given: Line $l_1$ passes through $A(2, 6, 2)$ and is normal to $2x + y - 2z = 10$ . Line $l_2$ is given by $\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$ .
goal: Find the shortest distance between skew lines $l_1$ and $l_2$ .
approach: The direction ratios of $l_1$ are along the normal to the plane: $\vec{p} = \langle 2, 1, -2 \rangle$ . Line $l_2$ passes through $B(-1, -4, 0)$ with direction $\vec{q} = \langle 2, -3, 2 \rangle$ . Shortest distance is $d = \frac{|(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$ .
execute: Compute $\vec{p} \times \vec{q}$ : $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(4 - (-4)) + \hat{k}(-6 - 2) = -4\hat{i} - 8\hat{j} - 8\hat{k}$ . Its magnitude is $|\vec{p} \times \vec{q}| = \sqrt{(-4)^2 + (-8)^2 + (-8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$ . Compute $\vec{a} - \vec{b} = \langle 2 - (-1), 6 - (-4), 2 - 0 \rangle = \langle 3, 10, 2 \rangle$ . Compute the scalar triple product: $(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q}) = 3(-4) + 10(-8) + 2(-8) = -12 - 80 - 16 = -108$ . Therefore, $d = \frac{|-108|}{12} = 9$ .
verify: Scalar determinant method directly: $\begin{vmatrix} 3 & 10 & 2 \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix} = 3(2 - 6) - 10(4 - (-4)) + 2(-6 - 2) = 3(-4) - 10(8) + 2(-8) = -12 - 80 - 16 = -108$ . Denominator is $|\langle -4, -8, -8 \rangle| = 12$ . Ratio is $108 / 12 = 9$ . Correct.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The shortest distance between the line $l_1$ and the given line is 9.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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