Q.Find the shortest distance between lines $\vec{r} = 6\hat{i} + 2\hat{j} + 2\hat{k} + \lambda (\hat{i} - 2\hat{j} + 2\hat{k})$ and $\vec{r} = -4\hat{i} - \hat{k} + \mu (3\hat{i} - 2\hat{j} - 2\hat{k})$.

The lines are skew, so the shortest distance is $d = \dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|}$.

Vectors. $\vec{a_1} = (6,2,2)$, $\vec{b_1} = (1,-2,2)$; $\vec{a_2} = (-4,0,-1)$, $\vec{b_2} = (3,-2,-2)$.

Cross product.

$\vec{b_1}\times\vec{b_2} = (8,\,8,\,4),\qquad |\vec{b_1}\times\vec{b_2}| = \sqrt{64+64+16} = 12.$ …

Why this formula

Two non-parallel lines that never meet are skew. The shortest gap between them is measured along the common perpendicular. The cross product $\vec{b_1}\times\vec{b_2}$ points along that common perpendicular direction, so projecting any join vector (from a point on one line to a point on the other) onto it gives the distance:

Read off the vectors

First line: $\vec{a_1} = (6,2,2)$, direction $\vec{b_1} = (1,-2,2)$.

Second line: $\vec{r} = -4\hat{i} - \hat{k} + \mu(3\hat{i}-2\hat{j}-2\hat{k})$, so $\vec{a_2} = (-4,0,-1)$, direction $\vec{b_2} = (3,-2,-2)$.

Cross product of the directions

• $\hat{i}$: $(-2)(-2)-(2)(-2) = 4+4 = 8$
• $\hat{j}$: $-\big[(1)(-2)-(2)(3)\big] = -(-2-6) = 8$
• $\hat{k}$: $(1)(-2)-(-2)(3) = -2+6 = 4$ …

Method: Shortest Distance Between Two Skew Lines

Use this when two lines in space are neither parallel nor intersecting and you need the shortest gap between them, measured along their common perpendicular.

Steps

Step 1: Read off a point and a direction from each line.

From $\vec{r}=\vec{a_1}+\lambda\vec{b_1}$ and $\vec{r}=\vec{a_2}+\mu\vec{b_2}$, identify the points $\vec{a_1},\vec{a_2}$ and directions $\vec{b_1},\vec{b_2}$. Watch for a missing component (e.g. no $\hat{j}$ term means that coordinate is $0$).

Step 2: Build the common-perpendicular direction and the join vector.

Compute $\vec{b_1}\times\vec{b_2}$ (perpendicular to both lines) and the join vector $\vec{a_2}-\vec{a_1}$.

Step 3: Project and divide. …

Common Mistakes

Mistake 1: Mis-reading the point of the second line when a component is missing.

Why it's wrong: $-4\hat{i}-\hat{k}$ has no $\hat{j}$ term, so its $y$-coordinate is $0$, giving $\vec{a_2}=(-4,0,-1)$; dropping or mis-placing that $0$ corrupts the join vector. Correct approach: write every point as an explicit $(x,y,z)$ triple, filling missing components with $0$.

Mistake 2: Forgetting the modulus in the numerator.

Why it's wrong: the scalar triple product came out $-108$, but a distance can never be negative. Correct approach: take $|-108|=108$, so $d=\tfrac{108}{12}=9$. …