Answer
$\dfrac{ \sqrt 2}{2}i -\dfrac{\sqrt 2}{2}j$
Work Step by Step
If a vector $v$ initiates at point $A(x_1,y_1)$ and terminates at $B(x_2,y_2)$, then
$v =\lt x_2-x_1, y_2-y_1 \gt =(x_2-x_1)i+(y_2-y_1) \ j$
Here, we have: $A=(3\sqrt 2, 7 \sqrt 2)$ and $B=(8\sqrt 2, 2 \sqrt 2)$
Therefore, $v=(8\sqrt 2-3 \sqrt 2)i+(2\sqrt 2-7 \sqrt 2) \ j \\= 5 \sqrt 2i -5 \sqrt 2 \ j$
The magnitude of a vector can be determined using the formula
$||v||=\sqrt{p^2+q^2} (1)$
We will use the formula (1) to obtain:
$||v||=\sqrt{(5\sqrt 2)^2+(-5 \sqrt 2)^2}\\
=\sqrt{50+50}\\
=\sqrt{100}\\=10$
The unit vector $u$ in the same direction as $v$ is given by: $u=\dfrac{v}{||v||}$
$u=\dfrac{v}{||v||}=\dfrac{5 \sqrt 2i -5 \sqrt 2 \ j}{10}=\dfrac{ \sqrt 2}{2}i -\dfrac{\sqrt 2}{2}j$
