Q.$\vec{AB} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{CD} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ are two vectors. The position vectors of the points $A$ and $C$ are $6\hat{i} + 7\hat{j} + 4\hat{k}$ and $-9\hat{j} + 2\hat{k}$, respectively. Find the position vector of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\vec{PQ}$ is perpendicular to $\vec{AB}$ and $\vec{CD}$ both.

Idea: Write $P$ and $Q$ as parametric points on the two lines, then impose $\vec{PQ}\cdot\vec{AB}=0$ and $\vec{PQ}\cdot\vec{CD}=0$.

Parametric points. With $A=6\hat{i}+7\hat{j}+4\hat{k}$, $\vec{AB}=3\hat{i}-\hat{j}+\hat{k}$, $C=-9\hat{j}+2\hat{k}$, $\vec{CD}=-3\hat{i}+2\hat{j}+4\hat{k}$:

Vector $\vec{PQ} = Q-P$:

$\vec{PQ} = (-3\lambda-3\mu-6)\hat{i} + (\lambda+2\mu-16)\hat{j} + (-\lambda+4\mu-2)\hat{k}.$ …

The idea

$P$ lies somewhere on line $AB$ and $Q$ somewhere on line $CD$, so each is described by a single parameter along its line. The segment $\vec{PQ}$ is the common perpendicular exactly when it is perpendicular to both direction vectors. That gives two dot-product equations in the two parameters $\lambda,\mu$ — enough to solve for both.

Set up

Work the steps

1. Form $\vec{PQ} = \vec{OQ}-\vec{OP}$.

which simplifies to

2. Perpendicular to $\vec{AB}=3\hat{i}-\hat{j}+\hat{k}$.

Expanding: $-9\lambda-9\mu-18-\lambda-2\mu+16-\lambda+4\mu-2 = -11\lambda-7\mu-4 = 0$, i.e.

3. Perpendicular to $\vec{CD}=-3\hat{i}+2\hat{j}+4\hat{k}$.

Expanding: $9\lambda+9\mu+18+2\lambda+4\mu-32-4\lambda+16\mu-8 = 7\lambda+29\mu-22 = 0$, i.e.

$7\lambda + 29\mu - 22 = 0. \quad (2)$ …

Method: Locating the endpoints of the common perpendicular of two lines

Use this when you must find the actual points $P$ and $Q$, one on each line, so that $\overrightarrow{PQ}$ is perpendicular to both lines (the feet of the shortest-distance segment).

Steps

Step 1: Parametrise each line separately.

Write $P=\vec a_1+\lambda\vec b_1$ and $Q=\vec a_2+\mu\vec b_2$ — two independent unknowns $\lambda,\mu$.

Step 2: Form the join.

Compute $\overrightarrow{PQ}=Q-P$ in terms of $\lambda$ and $\mu$.

Step 3: Impose both perpendicularities.

The two conditions …

Common Mistakes

Mistake 1: Using one parameter for both points.

Why it's wrong: $P$ and $Q$ move independently along different lines, so they need two separate parameters $\lambda$ and $\mu$. Correct approach: $P=\vec a_1+\lambda\vec b_1$, $Q=\vec a_2+\mu\vec b_2$, then solve two equations.

Mistake 2: Imposing perpendicularity to only one line.

Why it's wrong: the common perpendicular must be perpendicular to both directions; one condition leaves the answer undetermined. Correct approach: set $\overrightarrow{PQ}\cdot\vec b_1=0$ and $\overrightarrow{PQ}\cdot\vec b_2=0$ together. …