# If $\langle x,u\rangle = \langle x, v\rangle = \langle x,w\rangle = 0$ then $x=0$

I'm given that \begin{align*} u &= (1, 2, 3)\\ v &= (2, -1, 1)\\ w &= (3, 1, 0) \end{align*}

And I'm asked to verify if $\langle x,u\rangle = \langle x, v\rangle = \langle x,w\rangle = 0$ then $x=0$

I'm not really sure where to start. Logically, isn't it obvious? I know the following: \begin{align*} \langle x, u\rangle &= x_{1} + 2x_{2} + 3x_{3}.\\ \langle x, v\rangle &= 2x_{1} - x_{2} + x_{3}\\ \langle x, w\rangle &= 3x_{1} + x_{2} \end{align*}

But does that get me any where? I feel like I should be thinking about linearly independence and the matrix:

$$\begin{bmatrix} 1 & 2 & 3\\ 2 & -1 & 1\\ 3 & 1 & 0 \end{bmatrix}$$

Could someone point me in the right direction?

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If $x$ is orthogonal to $u$, $v$, and $w$, then it is orthogonal to the span of $u$, $v$, and $w$.
A vector in $\mathbb{R}^3$ is $\mathbf{0}$ if and only if it is orthogonal to every vector.
So, the conclusion will follow if and only if $u$, $v$, and $w$ span $\mathbb{R}^3$. if and only if $u$, $v$, and $w$ are linearly independent.
This amounts to the same thing you are doing: the system of linear equations you found has a nontrivial solution if and only if the determinant of your matrix is zero, if and only if the three vectors are not linearly independent. So if the vectors are not linearly independent, a nontrivial solution to the system of linear equations (a nonzero element in the nullspace of the matrix) will exhibit an $x$ that is not $\mathbf{0}$ but is orthogonal to each of $u$, $v$, and $w$. If the three vectors are linearly independent, then the only solution to the system is $\mathbf{x}=(0,0,0)$, which shows that the desired implication holds.