JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Shortest distance between two skew lines in 3D

Q: 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 ⟨ A, B, C ⟩.

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 (x+1)/2 = (y+4)/(-3) = z/2 is:

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A13/3B19/3C7D9

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 · 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 $b_{1} \times b_{2} = (- 4, - 8, - 8)$ with magnitude $\sqrt{16 + 64 + 64} = 12$ , and the vector connecting points is $(- 3, - 10, - 2)$ .

B · 19/3

Dividing the scalar triple product by the wrong vector magnitude or making an arithmetic error when evaluating $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2})$ . 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 $a_{2} - a_{1} = (- 3, - 10, - 2)$ , and the cross product of direction vectors is $(- 4, - 8, - 8)$ with magnitude 12, yielding a distance of $108 / 12 = 9$ . The shortest distance formula $d = \frac{∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣}{∣ b_{1} \times b_{2} ∣}$ was evaluated correctly.

Reviewed route

Solution

StepWorking

01Given

Line $l_{1}$ passes through $A (2, 6, 2)$ and is normal to $2 x + y - 2 z = 10$ . 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: $\vec{p} = ⟨ 2, 1, - 2 ⟩$ . Line $l_{2}$ passes through $B (- 1, - 4, 0)$ with direction $\vec{q} = ⟨ 2, - 3, 2 ⟩$ . Shortest distance is $d = \frac{∣ (\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q}) ∣}{∣ \vec{p} \times \vec{q} ∣}$ .

04Execute

Compute $\vec{p} \times \vec{q}$ : $\vec{p} \times \vec{q} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 2 & - 3 & 2 \end{array} ∣ = \hat{i} (2 - 6) - \hat{j} (4 - (- 4)) + \hat{k} (- 6 - 2) = - 4 \hat{i} - 8 \hat{j} - 8 \hat{k}$ . Its magnitude is $∣ \vec{p} \times \vec{q} ∣ = \sqrt{(- 4)^{2} + (- 8)^{2} + (- 8)^{2}} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$ . Compute $\vec{a} - \vec{b} = ⟨ 2 - (- 1), 6 - (- 4), 2 - 0 ⟩ = ⟨ 3, 10, 2 ⟩$ . 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: $∣ \begin{array}{ccc} 3 & 10 & 2 \\ 2 & 1 & - 2 \\ 2 & - 3 & 2 \end{array} ∣ = 3 (2 - 6) - 10 (4 - (- 4)) + 2 (- 6 - 2) = 3 (- 4) - 10 (8) + 2 (- 8) = - 12 - 80 - 16 = - 108$ . Denominator is $∣ ⟨ - 4, - 8, - 8 ⟩ ∣ = 12$ . Ratio is $108 / 12 = 9$ . Correct.

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $A x + B y + C z = D$ has its direction parallel to the normal vector of the plane, which is $⟨ A, B, C ⟩$ .

Answer

The shortest distance between the line $l_{1}$ and the given line is 9.

Why each option works or fails

A: 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 $b_{1} \times b_{2} = (- 4, - 8, - 8)$ with magnitude $\sqrt{16 + 64 + 64} = 12$ , and the vector connecting points is $(- 3, - 10, - 2)$ .
B: 19/3 - Dividing the scalar triple product by the wrong vector magnitude or making an arithmetic error when evaluating $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2})$ . 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 $a_{2} - a_{1} = (- 3, - 10, - 2)$ , and the cross product of direction vectors is $(- 4, - 8, - 8)$ with magnitude 12, yielding a distance of $108 / 12 = 9$ . The shortest distance formula $d = \frac{∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣}{∣ b_{1} \times b_{2} ∣}$ was evaluated correctly.

Step-by-step solution

Given: Line $l_{1}$ passes through $A (2, 6, 2)$ and is normal to $2 x + y - 2 z = 10$ . 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: $\vec{p} = ⟨ 2, 1, - 2 ⟩$ . Line $l_{2}$ passes through $B (- 1, - 4, 0)$ with direction $\vec{q} = ⟨ 2, - 3, 2 ⟩$ . Shortest distance is $d = \frac{∣ (\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q}) ∣}{∣ \vec{p} \times \vec{q} ∣}$ .
Execute: Compute $\vec{p} \times \vec{q}$ : $\vec{p} \times \vec{q} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 2 & - 3 & 2 \end{array} ∣ = \hat{i} (2 - 6) - \hat{j} (4 - (- 4)) + \hat{k} (- 6 - 2) = - 4 \hat{i} - 8 \hat{j} - 8 \hat{k}$ . Its magnitude is $∣ \vec{p} \times \vec{q} ∣ = \sqrt{(- 4)^{2} + (- 8)^{2} + (- 8)^{2}} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$ . Compute $\vec{a} - \vec{b} = ⟨ 2 - (- 1), 6 - (- 4), 2 - 0 ⟩ = ⟨ 3, 10, 2 ⟩$ . 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: $∣ \begin{array}{ccc} 3 & 10 & 2 \\ 2 & 1 & - 2 \\ 2 & - 3 & 2 \end{array} ∣ = 3 (2 - 6) - 10 (4 - (- 4)) + 2 (- 6 - 2) = 3 (- 4) - 10 (8) + 2 (- 8) = - 12 - 80 - 16 = - 108$ . Denominator is $∣ ⟨ - 4, - 8, - 8 ⟩ ∣ = 12$ . Ratio is $108 / 12 = 9$ . Correct.

Shortest distance between two skew lines in 3D

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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