3D Geometry: Common Mistakes and Fixes (3)

hard • 3 min read
Tags 3d Geometry

Question

Find the shortest distance between the lines

L1:x−12=y−23=z−34andL2:x−23=y−44=z−55L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \quad\text{and}\quad L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}

Then check whether these lines are skew, parallel, or intersecting.

Solution — Step by Step

L1L_1: passes through a⃗1=(1,2,3)\vec a_1 = (1, 2, 3), 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), direction b⃗2=(3,4,5)\vec b_2 = (3, 4, 5).

 b⃗1×b⃗2=∣i⃗j⃗k⃗234345∣=i⃗(15−16)−j⃗(10−12)+k⃗(8−9)\vec b_1 \times \vec b_2 = \begin{vmatrix} \vec i & \vec j & \vec k \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \vec i(15 - 16) - \vec j(10 - 12) + \vec k(8 - 9) b⃗1×b⃗2=(−1,2,−1)\vec b_1 \times \vec b_2 = (-1, 2, -1)

Magnitude: ∣b⃗1×b⃗2∣=1+4+1=6|\vec b_1 \times \vec b_2| = \sqrt{1 + 4 + 1} = \sqrt{6}.

a⃗2−a⃗1=(1,2,2)\vec a_2 - \vec a_1 = (1, 2, 2).

 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)(−1)+(2)(2)+(2)(−1)=−1+4−2=1(1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1.

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

Since d≠0d \ne 0 and b⃗1,b⃗2\vec b_1, \vec b_2 are not parallel (their cross product is nonzero), the lines are skew.

Shortest distance =16= \dfrac{1}{\sqrt{6}} units; lines are skew.

Why This Works

The cross product b⃗1×b⃗2\vec b_1 \times \vec b_2 is perpendicular to both lines. The shortest distance between two skew lines is the projection of any vector connecting one line to the other onto this common perpendicular — that is exactly what the formula computes.

If d=0d = 0, the lines either intersect or coincide, and we then check direction proportionality. If b⃗1×b⃗2=0⃗\vec b_1 \times \vec b_2 = \vec 0, the lines are parallel.

Alternative Method

Find the foot of the common perpendicular by parameterising both lines, setting up two scalar conditions (perpendicularity), and solving. Slower but conceptually cleaner — useful when the question asks for the foot, not just the distance.

Common Mistake

Using ∣a⃗2−a⃗1∣|\vec a_2 - \vec a_1| as the shortest distance. That gives the distance between two specific points on the lines, not the minimum distance. The cross-product formula projects the connecting vector onto the only direction that can give a true minimum.

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