Consider $ u, v\in\mathbb R^{ 3}$ and the following conditions : $||u − 3v|| = 5$, $||u|| = || − 2v|| = 2$, then prove $(u,v) = −2$
So this is what i did: Sorry if i made a mistake i am new with scalar product.
From $$||u - 3v||=5$$ i know : $$(u,u) + (u,-3v) + (-3v,u) + (-3v,-3v) = 5$$ But this is equal to: $$||u||^{ 2} - 6(u,v) - 9||v||^{ 2}= 5$$ From the conditions i know $||u||=||-2v||= |-2|||v||=2$, thus $||v||=1$
Now i know:
$$2^{ 2} - 6(u,v) - 9(1)^{ 2} = 5$$ thus:
$$(u,v)=\frac{-10}{ 6}$$
But this is not the result i wanted to prove. Any hint on what am i doing wrong?
$$(u,u)-6(u,v)+9(v,v)=25$$ or $$4-6(u,v)+9=25,$$ which gives $$(u,v)=-2.$$