## Calculus: Early Transcendentals 8th Edition

$u.v=\frac{1}{2}$ $u.w=-\frac{1}{2}$
The given shape is an equilateral triangle, thus, $|u|=1$ so $|v|=1$ Here, $\theta=\frac{\pi}{3}$ As we know $u.v=|u||v|cos\theta$ $u.v=|1||1|cos\frac{\pi}{3}=\frac{1}{2}$ Now, $u.w=|u||w|cos\theta$ The given shape is an equilateral triangle, thus, $|u|=1$ so $|w|=1$ Here, $\theta=\frac{2\pi}{3}$, because the vectors can only be found when the bases of the vectors are touching each other. If we move vector $u$ so its base is at the base of vector $w$, we see that $\theta=\frac{2\pi}{3}$ As we know $u.w=|u||w|cos\theta$ $u.w=|1||1|cos\frac{2\pi}{3}=-\frac{1}{2}$ Hence, $u.v=\frac{1}{2}$ $u.w=-\frac{1}{2}$