Computing the angle between two vectors using the inner product Question is: Consider $\mathbb{R}^3$ with the inner product $\langle u1,u2\rangle=x_1x_2+3y_1y_2+z_1z_2$ where $u_1=(x_1,y_1,z_1)$ and $u_2=(x_2,y_2,z_2)$ are two vectors in $\mathbb{R}^3$. What is the cosine of the angle between the vectors $v=(2,1,3)$ and $w=(-1,2,-1)$ with respect to the above inner product?

I don't even understand what exactly the inner product is,I tried using the formula
 $\cos(\theta)=\frac{\langle v,w\rangle}{|v||w|}$ but I don't understand how to use the inner product in this case. Any help would be much appreciated. I don't understand what to even substitute in the formula.
Hint: $$\cos\theta=\dfrac{\langle u_1,u_2\rangle}{|u_1||u_2|}\\
\implies \cos\theta=\dfrac{-2+6-3}{\sqrt{4+3+9}\sqrt{1+12+1}}$$
For the meaning of the inner product, see the images and the information here.
Hint: $$\cos\theta=\dfrac{\langle u_1,u_2\rangle}{|u_1||u_2|}\\
\implies \cos\theta=\dfrac{-2+6-3}{\sqrt{4+3+9}\sqrt{1+12+1}}$$
For the meaning of the inner product, see the images and the information here.