Q.Find the unit vector in the direction of vector $\vec{PQ}$, where P and Q are the points $(1, 2, 3)$ and $(4, 5, 6)$, respectively.

Concept: Direction Vectors — the vector from $P$ to $Q$ is $\vec{PQ} = \vec{Q} - \vec{P}$; a unit vector in that direction is $\frac{\vec{PQ}}{|\vec{PQ}|}$.

First, compute $\vec{PQ}$:

Next, find its magnitude:

The unit vector is: …

Why Direction Vectors Work

A vector between two points tells us two things: which way it points and how long it is. When we want only the direction — stripped of any length — we divide the vector by its own magnitude. That's the unit vector: a pure direction with length exactly 1.

For points $P(1,2,3)$ and $Q(4,5,6)$, the vector $\vec{PQ}$ runs from P to Q. Its components are simply the differences in each coordinate.

Step-by-step

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

   Subtract the coordinates of P from Q:

   $\vec{PQ} = (4-1,\ 5-2,\ 6-3) = (3,3,3)$

2. Compute its magnitude

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

   $|\vec{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}$

3. Divide the vector by its magnitude

   The unit vector $\hat{u}$ in the same direction is:

   $\hat{u} = \frac{\vec{PQ}}{|\vec{PQ}|} = \frac{(3,3,3)}{3\sqrt{3}} = \left(\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}\right)$

Method: Unit Vector Along a Directed Segment

Use this when you are given two points $P$ and $Q$ and asked for the unit vector along $\vec{PQ}$. It combines "displacement vector between two points" with "normalise a vector".

Steps

Step 1: Form the displacement vector, tip minus tail.

Order matters: for $\vec{PQ}$ you subtract the start $P$ from the end $Q$. Reversing this reverses the direction.

Step 2: Compute its magnitude.

$|\vec{PQ}|=\sqrt{(x_Q-x_P)^2+(y_Q-y_P)^2+(z_Q-z_P)^2}$ …

Common Mistakes

Mistake 1: Subtracting in the wrong order.

Why it's wrong: $\vec{PQ}=\vec{Q}-\vec{P}$, not $\vec{P}-\vec{Q}$. Using $\vec{P}-\vec{Q}$ gives $-\vec{PQ}$ — a unit vector pointing from $Q$ to $P$, the opposite direction to the one asked. Correct approach: always compute (end point) $-$ (start point).

Mistake 2: Mishandling $\sqrt{27}$.

Why it's wrong: $|\vec{PQ}|=\sqrt{27}=3\sqrt{3}$, not $27$, $9$, or $\sqrt{9}$. Correct approach: simplify the surd by taking out the largest perfect-square factor ($27=9\times3$). …