3D Geometry: Conceptual Doubts Cleared (2)

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Tags 3d Geometry

Question

Find the shortest distance between the lines r⃗1=i^+2j^+k^+λ(i^−j^+k^)\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k} + \lambda(\hat{i} - \hat{j} + \hat{k}) and r⃗2=2i^−j^−k^+μ(2i^+j^+2k^)\vec{r}_2 = 2\hat{i} - \hat{j} - \hat{k} + \mu(2\hat{i} + \hat{j} + 2\hat{k}).

Solution — Step by Step

For line 1: a⃗1=(1,2,1)\vec{a}_1 = (1, 2, 1) and direction b⃗1=(1,−1,1)\vec{b}_1 = (1, -1, 1). For line 2: a⃗2=(2,−1,−1)\vec{a}_2 = (2, -1, -1) and direction b⃗2=(2,1,2)\vec{b}_2 = (2, 1, 2).

 b⃗1×b⃗2=∣i^j^k^1−11212∣=i^(−2−1)−j^(2−2)+k^(1+2)=(−3,0,3)\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 + 2) = (-3, 0, 3)

Magnitude: ∣b⃗1×b⃗2∣=9+0+9=18=32|\vec{b}_1 \times \vec{b}_2| = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2}.

 a⃗2−a⃗1=(2−1,−1−2,−1−1)=(1,−3,−2)\vec{a}_2 - \vec{a}_1 = (2-1, -1-2, -1-1) = (1, -3, -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|}

Numerator: (1)(−3)+(−3)(0)+(−2)(3)=−3−6=−9(1)(-3) + (-3)(0) + (-2)(3) = -3 - 6 = -9. So ∣−9∣=9|{-9}| = 9.

d=932=32=322d = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}

Final Answer: d=322d = \dfrac{3\sqrt{2}}{2} units.

Why This Works

Two skew lines in 3D do not intersect and are not parallel. The shortest distance between them is along the common perpendicular, whose direction is given by b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2. The formula projects the connecting vector a⃗2−a⃗1\vec{a}_2 - \vec{a}_1 onto this perpendicular direction.

If the two lines were parallel, b⃗1×b⃗2=0⃗\vec{b}_1 \times \vec{b}_2 = \vec{0} and this formula breaks. For parallel lines, use the distance-from-point-to-line formula instead.

Alternative Method

Set up the perpendicularity conditions: the vector joining a general point on line 1 to a general point on line 2 must be perpendicular to both b⃗1\vec{b}_1 and b⃗2\vec{b}_2. Solve the resulting two-equation system in λ,μ\lambda, \mu to find the foot points, then compute the distance directly. Slower but conceptually cleaner.

Forgetting the absolute value in the numerator gives a negative answer, which makes no sense for a distance. Always wrap the dot product in ∣⋅∣|\cdot|.

If the scalar triple product (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0, the lines are coplanar (intersecting or parallel). This is a one-line check before computing the distance.

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