Prove that $\vec{u}$ and $\vec{v}$ are orthogonal vectors

Question

The vectors $\vec{u}$ and $\vec{v}$ are given in terms of the basis vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ as follows:

$\vec{u} = 3\vec{a} + 3\vec{b} - \vec{c}$

$\vec{v} = \vec{a} + 2\vec{b} + 3\vec{c}$

I've tried $\vec{u}.\vec{v}$ to see if their dot product equals to 0, but it does not. Am I missing something?

It was given that $\vec{a}, \vec{b},$ and $\vec{c}$ form a basis in $R^3$.

It was also given that:

$|\vec{a}| = 1, |\vec{b}| = 2,|\vec{c}| = 3$

Answer 1

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

Answer 2

My guess is that there is typo regarding signs in $u$ and $v$

We assume an ortonormal basis .

The way we have it now the dot product is $$ 3|a|^2+6|b|^2-3|c|^2=6$$

If we change $u$ to $-3a+3b -c$ then we have orthogonality.

Answer 3

If you expand $u\cdot v$ you find $9a\cdot b+8 a\cdot c+7b\cdot c$.
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.