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Question

The shortest distance between the lines

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

and

\dfrac{x-2}{3} = \dfrac{y-4}{4} = \dfrac{z-5}{5}

is:

(A)

\dfrac{1}{\sqrt{6}}

(B)

\dfrac{1}{6}

(C)

\dfrac{1}{\sqrt{3}}

(D)

\sqrt{6}

Solution Path

Skew lines distance via cross product (-1,2,-1), magnitude sqrt(6), scalar triple product = 1. Distance = 1/sqrt(6).

01Question Setup

1/4

Find the shortest distance between

L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}

and

L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}

d = \;?

02Cross Product

2/4

Direction vectors

\vec{b_1}=(2,3,4)

\vec{b_2}=(3,4,5)

. Cross product

\vec{b_1} \times \vec{b_2} = -\hat{i}+2\hat{j}-\hat{k}

\vec{b_1} \times \vec{b_2} = (-1, 2, -1)

03Scalar Triple ProductKEY INSIGHT

3/4

|\vec{b_1} \times \vec{b_2}| = \sqrt{6}

\vec{a_2}-\vec{a_1} = (1,2,2)

. Dot product

= -1+4-2 = 1

d = \frac{1}{\sqrt{6}}

04Final Answer

4/4

Shortest distance

= \frac{1}{\sqrt{6}}

. Verified: numerator

= |1| = 1

, denominator

= \sqrt{6}

\boxed{\frac{1}{\sqrt{6}}}

Concepts from this question2 concepts unlocked

★

Shortest Distance Between Skew Lines

TRICKY

Two lines in 3D that are neither parallel nor intersecting are called skew lines. The shortest distance between them is the length of the common perpendicular. For lines r = a1 + t*b1 and r = a2 + s*b2, the shortest distance is |(a2-a1).(b1 x b2)| / |b1 x b2|.

d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}

Shortest distance between skew lines is a high-weightage topic in JEE. It combines cross products, dot products, and scalar triple products in a single formula. Many students lose marks by mixing up the position and direction vectors.

Shortest distance between linesCoplanarity of linesSkew line problems

Practice (7 Qs) →

Equation of a Line in 3D

STANDARD

A line in 3D through point (x1, y1, z1) with direction ratios (a, b, c) is written in symmetric form as (x-x1)/a = (y-y1)/b = (z-z1)/c. In vector form: r = a + t*b, where a is the position vector of the point and b is the direction vector.

\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = t

The line equation is used in nearly every 3D geometry problem. Writing it correctly is the first step for finding intersections, shortest distances, and foot of perpendicular. JEE tests both the symmetric and vector forms.

Line through two pointsIntersection with planeShortest distanceFoot of perpendicular on a line

Practice (12 Qs) →

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