I am new to geometry and I can't solve this exercise..
I want to determine the distance between a point $ \omega= (0,0,0,0) \in \mathbb{R}^4 $
to a hyperplane $H:a_1x_1+a_2x_2+a_3x_3+a_4x_4=b$
I have a formula for the distance between a point $\omega$ to a hyperplane $H$ given by
$$|| v_0- \omega ||= \frac{ |b-( \omega,a) |}{ ||a||}$$
where $v_0$ is the orthogonal projection
Might one show me how to calculate this right ?
I could not find any examples which helped me. Thank you very much
Best Answer
Here's one way to do it. Let $\mathbf{x}$ be the vector from the origin achieving the minimal distance to the hyperplane having normal vector $\mathbf{a}$. Since $\mathbf{x}$ lies on the hyperplane we know (1) $\mathbf{a} \cdot \mathbf{x} = b$. Moreover, we know that (2) $\mathbf{a}$ and $\mathbf{x}$ are parallel. Note we can rewrite (2) as $|\mathbf{a} \cdot \mathbf{x}| = \|\mathbf{a}\| \| \mathbf{x}\| |\cos(\theta)| = \|\mathbf{a}\| \| \mathbf{x}\| \cdot 1$. Putting (1) and (2) together, we get $$|b| = \|\mathbf{a}\| \|\mathbf{x}\|.$$ In other words, $\|\mathbf{x}\| = |b|/\|\mathbf{a}\|$.