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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.

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    $\begingroup$ Please make titles informative!!!!!! $\endgroup$
    – Pedro Tamaroff
    Mar 31 '14 at 19:15
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    $\begingroup$ Does the angle change just because you are using another inner product? $\endgroup$
    – robjohn
    Mar 31 '14 at 19:15
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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.

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    $\begingroup$ The question is not using the standard inner product. $\endgroup$ Mar 31 '14 at 19:35
  • $\begingroup$ answer is 1/4sqrt(14) $\endgroup$
    – user134785
    Mar 31 '14 at 19:38
  • $\begingroup$ @SantiagoCanez I edited my answer. $\endgroup$
    – user122283
    Mar 31 '14 at 19:38
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    $\begingroup$ Still incorrect. The lengths depend on the inner product. $\endgroup$ Mar 31 '14 at 19:38
  • $\begingroup$ @SantiagoCanez so what exactly do we do? $\endgroup$
    – user134785
    Mar 31 '14 at 19:40

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