Find the distance and unit vector from P to Q

Key concept: This is a standard vector-from-points problem. First find the displacement vector $\overrightarrow{PQ}$, then use its length to build the unit vector.

Step by step

The position vectors are

$\overrightarrow{OP}=2\mathbf i-2\mathbf j=(2,-2),$ $\overrightarrow{OQ}=3\mathbf i-7\mathbf j=(3,-7).$

So

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=(3-2,\,-7-(-2))=(1,-5).$

a) Find $|\overrightarrow{PQ}|$

$|\overrightarrow{PQ}|=\sqrt{1^2+(-5)^2}=\sqrt{26}.$

b) Find the unit vector in the direction of $\overrightarrow{PQ}$

A unit vector is the vector divided by its magnitude:

$\hat{u}=\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{1}{\sqrt{26}}(1,-5).$

So the unit vector is

$\boxed{\frac{1}{\sqrt{26}}\mathbf i-\frac{5}{\sqrt{26}}\mathbf j}.$

Pitfall alert

Be careful with subtraction order: $\overrightarrow{PQ}$ means from P to Q, so it is $Q-P$, not the other way around. If you reverse it, the vector points in the opposite direction and the unit vector changes sign.

Try different conditions

If the question asked for $\overrightarrow{QP}$ instead, the magnitude would stay the same at $\sqrt{26}$, but the unit vector would become

$\frac{1}{\sqrt{26}}(-1,5).$

That sign change is easy to miss.

Further reading

displacement vector, magnitude, unit vector