Using cross product prove that if $\vec{u} \times \vec{v} = \vec{0}$ and $\vec{u} \cdot \vec{v} = 0$ then $\vec{v} = \vec{0}$

Question

I am asked to elaborate on the following proof:

Let $\vec{u} \neq \vec{0}$. Prove that if $\vec{u} \times \vec{v} = \vec{0}$ and $\vec{u} \cdot \vec{v} = 0$ then $\vec{v} = \vec{0}$.

My attempt was to say that

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

and that

$$ \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = 0 \Rightarrow (u_2 v_3 - v_2 u_3 , u_3 v_1 - u_1 v_3 , u_1 v_2 - u_2 v_1) = (0,0,0) $$

but it seems like the wrong way to go.

How can I achieve that proof?

Answer 1

Hint: Recall that $\vec{u}\times\vec{v}=0$ iff two non zero vectors are parallel, i.e. $ \vec{u}=\lambda \vec{v}.$ Conclude.

Answer 2

Hint: both $\sin$ and $\cos$ can't be $0$ at the same time. Note the formula for dot and cross product in terms of angles and magnitudes.

$$a\cdot b=\|a\|\|b\|\cos \theta$$

$$\|a\times b\|=\|a\|\|b\| \sin \theta $$