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More from Three Dimensional Geometry
MathMedMain 2026 • 4th Apr - Shift 2

Shortest Distance Between Two Lines in 3D Space

The shortest distance between the lines r⃗=(13i^+2j^+83k^)+λ(2i^−5j^+6k^)\vec{r} = \left(\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}\right) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k}) and r⃗=(−23i^−13k^)+μ(j^−k^),λ,μ∈R\vec{r} = \left(-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}\right) + \mu(\hat{j} - \hat{k}), \quad \lambda, \mu \in \mathbb{R}, is:

Options

A

\sqrt{5}

B

3

Correct
C

2\sqrt{3}

D

\sqrt{15}

Topics & Concepts

Three Dimensional GeometryLine in 3D

Step-by-Step Solution

To find the shortest distance between the two given lines in 3D space, we use the vector form of the lines: r⃗1=a⃗1+λb⃗1\vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1 r⃗2=a⃗2+μb⃗2\vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2

From the given equations of the lines: a⃗1=13i^+2j^+83k^,b⃗1=2i^−5j^+6k^\vec{a}_1 = \frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}, \quad \vec{b}_1 = 2\hat{i} - 5\hat{j} + 6\hat{k} a⃗2=−23i^−13k^,b⃗2=j^−k^\vec{a}_2 = -\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}, \quad \vec{b}_2 = \hat{j} - \hat{k}

The formula for the shortest distance dd between two skew lines is given by: 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=(−23i^−13k^)−(13i^+2j^+83k^)\vec{a}_2 - \vec{a}_1 = \left(-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}\right) - \left(\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}\right) a⃗2−a⃗1=(−23−13)i^−2j^+(−13−83)k^\vec{a}_2 - \vec{a}_1 = \left(-\frac{2}{3} - \frac{1}{3}\right)\hat{i} - 2\hat{j} + \left(-\frac{1}{3} - \frac{8}{3}\right)\hat{k} a⃗2−a⃗1=−i^−2j^−3k^\vec{a}_2 - \vec{a}_1 = -\hat{i} - 2\hat{j} - 3\hat{k}

Step 2: Calculate b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2 b⃗1×b⃗2=∣i^j^k^2−5601−1∣\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -5 & 6 \\ 0 & 1 & -1 \end{vmatrix} b⃗1×b⃗2=i^[(−5)(−1)−(6)(1)]−j^[(2)(−1)−(6)(0)]+k^[(2)(1)−(−5)(0)]\vec{b}_1 \times \vec{b}_2 = \hat{i}[(-5)(-1) - (6)(1)] - \hat{j}[(2)(-1) - (6)(0)] + \hat{k}[(2)(1) - (-5)(0)] b⃗1×b⃗2=i^(5−6)−j^(−2−0)+k^(2−0)\vec{b}_1 \times \vec{b}_2 = \hat{i}(5 - 6) - \hat{j}(-2 - 0) + \hat{k}(2 - 0) b⃗1×b⃗2=−i^+2j^+2k^\vec{b}_1 \times \vec{b}_2 = -\hat{i} + 2\hat{j} + 2\hat{k}

Step 3: Calculate the magnitude ∣b⃗1×b⃗2∣|\vec{b}_1 \times \vec{b}_2| ∣b⃗1×b⃗2∣=(−1)2+22+22=1+4+4=9=3|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3

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)=(−i^−2j^−3k^)⋅(−i^+2j^+2k^)(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (-\hat{i} + 2\hat{j} + 2\hat{k}) =(−1)(−1)+(−2)(2)+(−3)(2)= (-1)(-1) + (-2)(2) + (-3)(2) =1−4−6=−9= 1 - 4 - 6 = -9

Step 5: Calculate the shortest distance dd d=∣−9∣3=93=3d = \frac{|-9|}{3} = \frac{9}{3} = 3

Hence, the shortest distance between the given lines is 33.

Correct Option: B

Shortest Distance Between Two Lines in 3D Space | Mathematics PYQ Solution - JEE Challenger