Q.Find a unit vector in the direction of $\vec{PQ}$, where P and Q have co-ordinates $(5, 0, 8)$ and $(3, 3, 2)$, respectively.

Concept: Direction Vectors — the vector $\vec{PQ}$ is found by subtracting coordinates of the tail from the head.

Step 1: Compute $\vec{PQ}$.

Step 2: Find its magnitude.

Step 3: Divide the vector by its magnitude to get the unit vector.

Why Direction Vectors Work

A vector like $\vec{PQ}$ tells us two things: which way it points and how long it is. When we want "a unit vector in the direction of $\vec{PQ}$", we're asking for a vector that points exactly the same way but has length exactly 1. That's just the original vector scaled down by its own length — like shrinking a rope to exactly one metre without changing its orientation.

The formula is simple: if $\vec{v}$ is any non-zero vector, the unit vector in its direction is $\frac{\vec{v}}{|\vec{v}|}$.

Step-by-step

1. Find the vector $\vec{PQ}$.

The vector from P to Q is obtained by subtracting the coordinates of P from those of Q:

So $\vec{PQ} = -2\hat{i} + 3\hat{j} - 6\hat{k}$.

2. Compute the magnitude (length) of $\vec{PQ}$.

The magnitude of a vector $(x, y, z)$ is $\sqrt{x^2 + y^2 + z^2}$:

3. Divide the vector by its magnitude.

The unit vector $\hat{u}$ in the direction of $\vec{PQ}$ is:

4. Verify the result.

A quick check: the magnitude of $\hat{u}$ should be 1.

It works.

Method: Unit vector along the segment joining two points

Use this when two points are given and you need a unit vector in the direction from one to the other.

Steps

Step 1: Build the displacement vector as head minus tail.

For $\vec{PQ}$ (from $P$ to $Q$),

subtracting coordinate by coordinate. The order matters: $\vec{PQ}$ starts at $P$, so subtract $P$.

Step 2: Compute the magnitude.

Step 3: Divide the displacement by its magnitude.

The components of $\hat{u}$ are exactly the direction cosines of the segment.

Common Mistakes

Mistake 1: Computing $\vec{PQ}$ as $P-Q$ instead of $Q-P$.

Why it's wrong: $\vec{PQ}$ runs from $P$ to $Q$, so it is $Q-P=(-2,3,-6)$; reversing it gives the opposite direction $\vec{QP}$. Correct approach: subtract the tail ($P$) from the head ($Q$).

Mistake 2: Reporting $\vec{PQ}$ itself as the answer.

Why it's wrong: the question asks for a unit vector, and $|\vec{PQ}|=7\ne1$. Correct approach: divide $\vec{PQ}$ by its magnitude $7$.

Mistake 3: Arithmetic slip in the magnitude.

Why it's wrong: $|\vec{PQ}|=\sqrt{4+9+36}=\sqrt{49}=7$; mis-squaring a negative coordinate (e.g. $(-6)^2=36$, not $-36$) breaks this. Correct approach: square each component to a non-negative value before summing.