Using A Specified Inner Product to Calculate Cos(θ)

Question

Hello I am getting stuck on this question:

What I have done to attempt to solve this problem is use u⋅v=|u||v|cosθ to solve for cosθ, but instead of standard dot product I was using the one specified in the question to calculate (u,v), (u,u) and (u,v) however I keep getting E while the answer is A. Any help is appreciated!

Answer 1

By definition, $$ \cos(\theta) =\frac{\langle u, v\rangle}{\lVert u\rVert\lVert w\rVert} $$ Taking $u=\langle-1,1\rangle$ and $v=\langle 2,1\rangle$, one easily computes \begin{align*} \langle u,v\rangle &= -9 & \lVert u\rVert &=\sqrt{\langle u, u\rangle} =\sqrt{9}=3 & \lVert v\rVert &=\sqrt{\langle v, v\rangle} =\sqrt{36}=6 \end{align*} Hence $$ \cos(\theta)=-\frac{9}{18}=-\frac{1}{2} $$

Answer 2

Let $u=(-1,1)$ and $v=(2,1)$. We will also write $M=\begin{bmatrix}7&1\\1&4\end{bmatrix}$. Then $Mu=(-6,3)$ and $u^\top Mu=6+3=9$. Thus, $|u|=\sqrt{9}=3$. We have $Mv=(15,6)$ and $v^\top Mv=36$ so $|v|=\sqrt{36}=6$. Finally, we have $u^\top Mv=-9$. Thus, $\langle u,v\rangle=-9=6\cdot 3\cos(\theta)$. Thus, $\cos(\theta)=\frac{-1}{2}$.

Answer 3

Your inner product is $$\langle u,v \rangle = u^T \begin{bmatrix} 7 & 1 \\1 & 4 \end{bmatrix} v,$$ so your new vector norm is $$| u| = \sqrt{\langle u, u \rangle} = \sqrt{u^T \begin{bmatrix} 7 & 1 \\1 & 4 \end{bmatrix} u}.$$ Then using the formula $$\langle u,v \rangle = |u| |v| \cos{\theta}$$ we have $$\cos{\theta}= \frac{\langle u,v \rangle}{|u| |v|} = \frac{u^T \begin{bmatrix} 7 & 1 \\1 & 4 \end{bmatrix} v}{\sqrt{u^T \begin{bmatrix} 7 & 1 \\1 & 4 \end{bmatrix} u} \sqrt{v^T \begin{bmatrix} 7 & 1 \\1 & 4 \end{bmatrix} v}} = \frac{-9}{(3)(6)} = -\frac{1}{2}.$$