3D Geometry: PYQ Walkthrough (4)

easy • 3 min read
Tags 3d Geometry

Question

(JEE Main 2024, Shift 1 — paraphrased) Find the shortest distance between the lines:

L1:x−12=y−23=z−34L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}

L2:x−23=y−44=z−55L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}

Solution — Step by Step

L1L_1 passes through a⃗1=(1,2,3)\vec{a}_1 = (1, 2, 3) with direction b⃗1=(2,3,4)\vec{b}_1 = (2, 3, 4).

L2L_2 passes through a⃗2=(2,4,5)\vec{a}_2 = (2, 4, 5) with direction b⃗2=(3,4,5)\vec{b}_2 = (3, 4, 5).

Vector joining the points: a⃗2−a⃗1=(1,2,2)\vec{a}_2 - \vec{a}_1 = (1, 2, 2).

b⃗1×b⃗2=∣i^j^k^234345∣\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}

=i^(3⋅5−4⋅4)−j^(2⋅5−4⋅3)+k^(2⋅4−3⋅3)= \hat{i}(3 \cdot 5 - 4 \cdot 4) - \hat{j}(2 \cdot 5 - 4 \cdot 3) + \hat{k}(2 \cdot 4 - 3 \cdot 3)

=i^(15−16)−j^(10−12)+k^(8−9)=(−1,2,−1)= \hat{i}(15 - 16) - \hat{j}(10 - 12) + \hat{k}(8 - 9) = (-1, 2, -1)

Magnitude: ∣b⃗1×b⃗2∣=1+4+1=6|\vec{b}_1 \times \vec{b}_2| = \sqrt{1 + 4 + 1} = \sqrt{6}.

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|}

Dot product: (1,2,2)⋅(−1,2,−1)=−1+4−2=1(1, 2, 2) \cdot (-1, 2, -1) = -1 + 4 - 2 = 1.

d=∣1∣6=16d = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}

Shortest distance = 16\frac{1}{\sqrt{6}} units.

Why This Works

The shortest distance between two skew lines lies along the common perpendicular. The direction of this perpendicular is b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2. Projecting the joining vector (a⃗2−a⃗1)(\vec{a}_2 - \vec{a}_1) onto this perpendicular direction gives the shortest distance directly.

If the lines were parallel, b⃗1×b⃗2=0⃗\vec{b}_1 \times \vec{b}_2 = \vec{0} and the formula breaks down — for parallel lines, use a different formula involving ∣b⃗1×(a⃗2−a⃗1)∣|\vec{b}_1 \times (\vec{a}_2 - \vec{a}_1)|.

If the lines intersect, d=0d = 0, which means (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0 — this is the coplanarity condition.

If a JEE question asks for the shortest distance and gives lines that “look” parallel (proportional direction ratios), check first — you’ll need a different formula.

Alternative Method

Parameterize both lines: P1=(1+2t,2+3t,3+4t)P_1 = (1+2t, 2+3t, 3+4t) and P2=(2+3s,4+4s,5+5s)P_2 = (2+3s, 4+4s, 5+5s). Set the vector P2P1⃗\vec{P_2 P_1} perpendicular to both b⃗1\vec{b}_1 and b⃗2\vec{b}_2. Solve the resulting two equations for t,st, s, then compute ∣P2P1⃗∣|\vec{P_2 P_1}|. Same answer, much longer — only worth it if you forget the formula.

Common Mistake

Students forget to take the absolute value of the dot product in the numerator. If you get a negative number, take its magnitude — distance is always non-negative.

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