Show that, if $u,v,w$ are orthogonal two-by-two, then $S = \{ u , v , w\}$ forms a basis which is linearly independent

Question

I am given the following question:

Show that, if $u,v,w$ are orthogonal two-by-two, then $S = \{ u , v , w\}$ forms a basis which is linearly independent.

My idea to tackle this problem is to say that

$$ u . v = 0 \Rightarrow (u_1, u_2, u_3) \cdot (v_1, v_2, v_3) = 0 \Rightarrow u_1 v_1 + u_2 v_2 + u_3 v_3 = 0\\ v . w = 0 \Rightarrow (v_1, v_2, v_3) \cdot (w_1, \cdots \\ \cdots $$

and that

$$ \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} \neq 0 $$

but I'm pretty sure that's not the way to do it. Can someone help me?

Answer 1

If $$au+bv+cw=0$$ Then dot product by $u$ gives $$a u\cdot u + 0 + 0 = 0$$ $$a=0$$ Dot product by $v$ gives $$0+b v \cdot v + 0=0$$ $$b=0$$ Dot product by $w$ gives $$0+0+c w\cdot w=0$$ $$c=0$$ Hence they are linearly independent.