How can I compare two vectors if their inner product is zero?

Question

I have the following:

$\vec{w} = \begin{bmatrix}1 \\ \\ 2\end{bmatrix},$ where $\vec{w}$ is a position vector.

If I do a vector multiplication with $\vec{w}$ and two coordinates $x_1$ and $x_2$,

$\begin{bmatrix}1 \\ \\ 2\end{bmatrix}^T \begin{bmatrix}x_1 \\ \\ x_2\end{bmatrix} = 3$, where $3$ is the intercept.

The above matrix equation becomes

$x_1 + 2x_2 - 3 = 0$

or

$x_2 = -\dfrac{1}{2}x_1 + \dfrac{3}{2}$

The graphical presentation of the relationship between $\vec{w}$ and the above equation $x_2 = -\dfrac{1}{2}x_1 + \dfrac{3}{2}$ becomes perpendicular as follows:

But how can I show the equation and $\vec{w}$ is perpendicular?

Answer 1

The line $OW$ has its equation $x_2=2x_1$ with slope $2$ and the other line has slope $\dfrac{-1}{2}$ so their product is $-1$.

Answer 2

One way is to calculate the $w$ which is closest from origo. All points $w$ closest to a differentiable manifold will be perpendicular to the surface tangent space (and parallell to the normal) thanks to a result in differential geometry. A line is maybe the simplest differentiable manifold that exists.

So we can solve $$w_O = \min_w \{\|w\|_2 + \|w_1+2w_2-3\|_2\}$$

Which is a linear least squares problem.