Nelson Jennings

2022-07-18

Prove that $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ are orthogonal vectors
The vectors $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ are given in terms of the basis vectors $\stackrel{\to }{a}$ , $\stackrel{\to }{b}$ and $\stackrel{\to }{c}$ as follows:
$\stackrel{\to }{u}=3\stackrel{\to }{a}+3\stackrel{\to }{b}-\stackrel{\to }{c}$
$\stackrel{\to }{v}=\stackrel{\to }{a}+2\stackrel{\to }{b}+3\stackrel{\to }{c}$
I've tried $\stackrel{\to }{u}.\stackrel{\to }{v}$ to see if their dot product equals to 0, but it does not. Am I missing something?
It was given that $\stackrel{\to }{a},\stackrel{\to }{b},$ and $\stackrel{\to }{c}$ form a basis in ${R}^{3}$.
It was also given that:
$|\stackrel{\to }{a}|=1,|\stackrel{\to }{b}|=2,|\stackrel{\to }{c}|=3$

Steppkelk

Expert

With these changes,
$⟨u,v⟩=⟨3a+3b-c,a+2b+3c⟩\stackrel{!}{=}⟨3a,a⟩+⟨3b,2b⟩+⟨-c,3c⟩=3⟨a,a⟩+6⟨b,b⟩-3⟨c,c⟩=3\cdot {1}^{2}+6\cdot {2}^{2}-3\cdot {3}^{2}=0$
But you still need that the basis is orthogonal.

makaunawal5

Expert

If you expand $u\cdot v$ you find $9a\cdot b+8a\cdot c+7b\cdot c$
There is no reason for this to be zero in general, unless we also know that a,b,c are orthogonal. So my guess is that in the instructions for the problem there was the information that a,b,c are orthogonal.

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