Show $\langle u,v \rangle = -\frac{1}{2}$ when $u+v+w=0$ and $\|u\|=\|v\|=\|w\|=1$?

Question

Show $\langle u,v \rangle = -\frac{1}{2}$ when $u+v+w=0$ and $\|u\|=\|v\|=\|w\|=1$?

My thinking is:

$\langle u+v+w,v \rangle =0 \iff \langle u,v \rangle + \langle w,v \rangle = -1$

How do i continue?

Answer 1

$$\langle w,w\rangle = 1$$ leads to $$\langle -u-v,-u-v\rangle =1.$$ Expand and simplify and you're done.

Answer 2

\begin{align*} \langle u+v+w, u+v-w\rangle&=0\\ \|u\|^2+\|v\|^2+2\langle u, v\rangle-\|w\|^2=0\\ 2\langle u, v\rangle+1&=0\\ \langle u, v\rangle&=-\frac{1}{2} \end{align*}

Answer 3

As others have shown, the algebraic method to prove the assertion in question isn't too hard but it occurred to me to think about a geometrical proof. Since the 3 vectors add to the zero vector they form a triangle. Since they have the same magnitude, that triangle is equilateral. An equilateral triangle has its 3 internal angles all equal to 60 degrees. The inner or dot product between any two of the vectors is the product of their magnitudes and the cosine of the angle between them. Their magnitudes are 1 and the angle between them is $180 - 60 = 120$ degrees, hence the dot product of any two of the 3 vectors equals $\cos(2\pi/3) = -1/2$.