Q.(a) Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are $\vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda(7\hat{i} - 6\hat{j} + \hat{k})$ and $\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu(\hat{i} - 2\hat{j} + \hat{k})$, where $\lambda$ and $\mu$ are parameters.

Concept: Distance Between Skew Lines — the shortest distance between two non-parallel, non-intersecting lines is the length of the common perpendicular.

Step 1 – Check if lines are parallel

Direction vectors: $\vec{d}_1 = 7\hat{i} - 6\hat{j} + \hat{k}$, $\vec{d}_2 = \hat{i} - 2\hat{j} + \hat{k}$.

Since $\vec{d}_1$ is not a scalar multiple of $\vec{d}_2$, the lines are skew (not parallel).

Step 2 – Use the distance formula

For lines $\vec{r} = \vec{a}_1 + \lambda \vec{d}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{d}_2$,

Here $\vec{a}_1 = -\hat{i} - \hat{j} - \hat{k}$, $\vec{a}_2 = 3\hat{i} + 5\hat{j} + 7\hat{k}$, so

$\vec{a}_2 - \vec{a}_1 = 4\hat{i} + 6\hat{j} + 8\hat{k}$.

Step 3 – Compute cross product

$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = (-6\cdot1 - 1\cdot(-2))\hat{i} - (7\cdot1 - 1\cdot1)\hat{j} + (7\cdot(-2) - (-6)\cdot1)\hat{k}$ …

Read off the vectors.

Direction cross product:

Connecting vector and its projection: …

Method: Shortest distance between two skew lines

Use this for the shortest (common-perpendicular) distance between two non-parallel, non-intersecting lines $\vec{r}=\vec{a}_1+\lambda\vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu\vec{b}_2$.

Steps

Step 1: Confirm the lines are skew (not parallel).

$\vec{b}_1$ and $\vec{b}_2$ must not be proportional, so $\vec{b}_1\times\vec{b}_2\neq\vec{0}$. (If they ARE parallel, this formula does not apply — use the parallel-line formula instead.)

Step 2: Apply the formula.

$d=\frac{\left|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)\right|}{\left|\vec{b}_1\times\vec{b}_2\right|}.$ …

Common Mistakes

Mistake 1: Forgetting to divide by $|\vec{b}_1\times\vec{b}_2|$.

Why it's wrong: the numerator $(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)$ is a scalar triple product (a volume), not a distance; dividing by $|\vec{b}_1\times\vec{b}_2|$ turns it into a length. Correct approach: $d=\frac{116}{2\sqrt{29}}=2\sqrt{29}$.

Mistake 2: Using the parallel-line distance formula. …