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MathMedMain 2026 • 6th Apr - Shift 2

Shortest Distance Between Two Lines in Space

The shortest distance between the lines x−41=y−32=z−2−3\frac{x - 4}{1} = \frac{y - 3}{2} = \frac{z - 2}{-3} and x+22=y−64=z−5−5\frac{x + 2}{2} = \frac{y - 6}{4} = \frac{z - 5}{-5} is :

Options

A

566\frac{5\sqrt{6}}{6}

B

252\sqrt{5}

C

353\sqrt{5}

Correct
D

454\sqrt{5}

Topics & Concepts

Three Dimensional GeometryLine in 3D

Step-by-Step Solution

To find the shortest distance between the two given lines in space, we first write their vector equations.

The equation of the first line L1L_1 is given by: x−41=y−32=z−2−3\frac{x - 4}{1} = \frac{y - 3}{2} = \frac{z - 2}{-3}

From this, line L1L_1 passes through the point with position vector a⃗1=4i^+3j^+2k^\vec{a}_1 = 4\hat{i} + 3\hat{j} + 2\hat{k} and is parallel to the vector b⃗1=i^+2j^−3k^\vec{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k}.

The equation of the second line L2L_2 is given by: x+22=y−64=z−5−5\frac{x + 2}{2} = \frac{y - 6}{4} = \frac{z - 5}{-5}

From this, line L2L_2 passes through the point with position vector a⃗2=−2i^+6j^+5k^\vec{a}_2 = -2\hat{i} + 6\hat{j} + 5\hat{k} and is parallel to the vector b⃗2=2i^+4j^−5k^\vec{b}_2 = 2\hat{i} + 4\hat{j} - 5\hat{k}.

The formula for the shortest distance dd between two non-parallel skew lines is: d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}

Step 1: Calculate a⃗2−a⃗1\vec{a}_2 - \vec{a}_1 a⃗2−a⃗1=(−2−4)i^+(6−3)j^+(5−2)k^=−6i^+3j^+3k^\vec{a}_2 - \vec{a}_1 = (-2 - 4)\hat{i} + (6 - 3)\hat{j} + (5 - 2)\hat{k} = -6\hat{i} + 3\hat{j} + 3\hat{k}

Step 2: Calculate the cross product b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2 b⃗1×b⃗2=∣i^j^k^12−324−5∣\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}

b⃗1×b⃗2=i^(2(−5)−(−3)(4))−j^(1(−5)−(−3)(2))+k^(1(4)−2(2))\vec{b}_1 \times \vec{b}_2 = \hat{i}(2(-5) - (-3)(4)) - \hat{j}(1(-5) - (-3)(2)) + \hat{k}(1(4) - 2(2)) b⃗1×b⃗2=i^(−10+12)−j^(−5+6)+k^(4−4)\vec{b}_1 \times \vec{b}_2 = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4) b⃗1×b⃗2=2i^−j^+0k^\vec{b}_1 \times \vec{b}_2 = 2\hat{i} - \hat{j} + 0\hat{k}

Step 3: Calculate the magnitude ∣b⃗1×b⃗2∣|\vec{b}_1 \times \vec{b}_2| ∣b⃗1×b⃗2∣=22+(−1)2+02=4+1=5|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5}

Step 4: Calculate the dot product (a⃗2−a⃗1)⋅(b⃗1×b⃗2)(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(−6i^+3j^+3k^)⋅(2i^−j^+0k^)(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-6\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (2\hat{i} - \hat{j} + 0\hat{k}) =(−6)(2)+(3)(−1)+(3)(0)=−12−3+0=−15= (-6)(2) + (3)(-1) + (3)(0) = -12 - 3 + 0 = -15

Step 5: Calculate the shortest distance dd d=∣−15∣5=155=35d = \frac{|-15|}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}

Thus, the shortest distance between the lines is 353\sqrt{5}.

Correct Option: C

Shortest Distance Between Two Lines in Space | Mathematics PYQ Solution - JEE Challenger