Show that vector v is equal to the zero vector

Let H = Span{$\vec v$₁, $\vec v$₂} where {$\vec v$₁, $\vec v$₂} is an orthogonal set of nonzero vectors. Let $\vec v$ ∈ H such that
$\vec v$ ⋅$\vec v$₁ = 0 and $\vec v$ ⋅ $\vec v$₂ = 0. Show that $\vec v$ = 0. Hint: Compute $||\vec v||^2 = \vec v ⋅ \vec v$ by writting $\vec v$ as a linear combination of $\vec v_{1}, \vec v_{2}$ and taking the dot product with $\vec v$.

Since I know that ||$\vec v$|| = 0 if and only if $\vec v$ = $\vec 0$ I found the weights like this: $$c_{1} = \frac{\vec v ⋅ \vec v_{1}}{\vec v_{1}⋅\vec v_{1}} =\frac{\vec 0 ⋅ \vec v_{1}}{\vec v_{1}⋅\vec v_{1}} = 0$$ $$c_{2} = \frac{\vec v ⋅ \vec v_{2}}{\vec v_{2}⋅\vec v_{2}} = \frac{\vec 0 ⋅ \vec v_{2}}{\vec v_{2}⋅\vec v_{2}} = 0$$ Using that to create a linear combination of $\vec v_{1},\vec v_{2}$

$$\vec v = 0\vec v_{1} + 0\vec v_{2}$$ $$\vec v = 0$$

Is this even correct? Or am i just overthinking the question?

Solution 1:

Following the given hint we have

$$v=av_1+bv_2 \implies v\cdot v=av\cdot v_1+bv\cdot v_2=0+0=0$$

and then we also have

$$v=av_1+bv_2 \implies v\cdot v= a^2|v_1|^2+b^2|v_2|^2=0$$

and the latter holds $\iff a=b=0$.