## Calculus: Early Transcendentals 8th Edition

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