Finding angle between vectors given dot products

Question

I am given the following information on the inner products ($\cdot$) between vectors $u, v, \text{and } w$ in $n$-dimensional Euclidean space: $u\cdot u=1, v \cdot v=8, w \cdot w=50, u \cdot v=0, u \cdot w=8, v \cdot w =3$.
I am trying to find the angle between $u$ and $v+w$. My idea is to use the dot product identity; I know that the angle $\theta$ satisfies $u \cdot (v+w)=|u||v+w|\cos\theta$, where $|\cdots|$ denotes the Euclidean norm. I know I can use the distributive property on the left-hand side, which simplifies to $8$, but how can I evaluate the other terms to solve for $\theta$?

Answer 1

So you want to find $||\vec u+ \vec v||$.

Let $\vec u=<u_1,u_2,...,u_n>$ and $\vec v=<v_1,v_2,v_3,...,v_n>$. Then $||\vec u+\vec v||=\sqrt{(u_1+v_1)^2+(u_2+v_2)^2...+(u_n+v_n)^2}$. Expand inside the square root. Can you rewrite after that?

Hints:

$\vec u \cdot \vec u=u_1^2+u_2^2...+u_n^2$

$\vec v \cdot \vec v=v_1^2+v_2^2...+v_n^2$

$\vec u \cdot \vec v=u_1v_1+u_2v_2...+u_nv_n$

To find $||u||$ just note that $\vec u \cdot \vec u=||u||^2$.

Answer 2

From $\vec u \cdot \vec u = 1 \implies |\vec u|^2 = 1 \implies |\vec u| = 1$

Now consider $(\vec v+\vec w) \cdot (\vec v+\vec w) = \vec v \cdot \vec v + 2\vec w \cdot \vec v + \vec w \cdot \vec w = |\vec v+\vec w|^2$

From these, you should be able to figure out what you need to evaluate that expression.