#### Answer

$u.v=\frac{1}{2}$ and
$u.w=0$

#### Work Step by Step

The given shape is an equilateral triangle, thus, $|u|=1$ and $|v| $ is a leg of a $45^\circ-45^\circ-90^\circ$ triangle where the hypotenuse is 1, we know it is $|u|=\frac{1}{\sqrt 2}$
Here, $\theta=45^\circ$
As we know $u.v=|u||v|cos\theta$
$u.v=|1||\frac{1}{\sqrt 2}|cos45^\circ$
$=1\times\frac{1}{\sqrt 2}\times\frac{1}{\sqrt 2}$
$=\frac{1}{2}$
We now that the dot product is always $0$ when two vectors are orthogonal and $u$ and $w$ are orthogonal in the given shape.
Therefore, $u.w=0$
Hence, $u.v=\frac{1}{2}$ and
$u.w=0$