TL;DR

Additional notes on the orthogonality of two vectors.

In previous post PWC123 - Square Points there is this statement, which might appear as having been drawn out of thin air:

Checking for orthogonality can be done calculating their regular scalar (or dot) product:

$v \cdot w = v_x w_x + v_y w_y$

This is 0 if and only if the two vectors are orthogonal, so itâ€™s exactly the condition we are after.

So we have two vectors $\vec{v} = (v_1, â€¦, v_n)$ and $\vec{w} = (w_1, â€¦, w_n)$ (where $n = 2$ in the case of the previous post, but weâ€™re aiming for the big, general case here) and we want to understand whether theyâ€™re orthogonal or not, based on their scalar product. Letâ€™s go!

If they are, then $\vec{v}$ is also orthogonal to $-\vec{w}$, because $\vec{w}$ and $-\vec{w}$ are 180Â° apart from each other.

Letâ€™s now consider a triangle $\overset{\triangle}{ABC}$ where:

$A = O + \vec{w} \\ B = O - \vec{w} \\ C = O + \vec{v}$

Sub-triangles $\overset{\triangle}{AOC}$ and $\overset{\triangle}{BOC}$ are congruent, which practically speaking means that theyâ€™re the same triangle with some translation and/or rotation and/or flipping (but no scaling). They satisfy the so-called Side-Angle-Side (SAS) condition for congruence, because:

• $\overline{OA}$ and $\overline{OB}$ have the same length:
$L_{\overline{OA}} = |\vec{w}| = |-\vec{w}| = L_{\overline{OB}}$
• $\overline{OC}$ is in common;

• angles in between $\widehat{AOC}$ and $\widehat{BOC}$ are both 90Â°.

This implies that segments $\overline{AC}$ and $\overline{BC}$ have the same length so, by extension, the square of their lengths:

$L_{\overline{AC}}^2 = |\vec{v} - \vec{w}|^2 = \sum_{i=1}^{n}(v_i - w_i)^2 = \sum_{i=1}^{n}(v_i^2 - 2 v_i w_i + w_i^2) \\ L_{\overline{BC}}^2 = |\vec{v} - (-\vec{w})|^2 = \sum_{i=1}^{n}(v_i + w_i)^2 = \sum_{i=1}^{n}(v_i^2 + 2 v_i w_i + w_i^2)$

are the same, i.e. their difference is $0$:

$L_{\overline{BC}}^2 - L_{\overline{AC}}^2 = 0 \\ \sum_{i=1}^{n}(v_i^2 + 2v_i w_i + w_i^2 - v_i^2 + 2v_i w_i - w_i^2) = 0 \\ 4 \sum_{i=1}^{n}v_i w_i = 0 \\ \sum_{i=1}^{n}v_i w_i = 0 \iff \vec{v} \cdot \vec{w} = 0$

i.e. the scalar product (a.k.a. dot product) between $\vec{v}$ and $\vec{w}$ is $0$.

On the other hand, itâ€™s easy to go in the opposite direction: if the scalar product is $0$, then either one of the two vectors is the null vector (which is assumed to be orthogonal to any other vector, by definition), or they must form a triangle that satisfy the relation above where the length of $\overline{AC}$ is the same as the length of $\overline{BC}$, which in turn implies that segment $\overline{OC}$ is the height of the isosceles triangle and that angle $\widehat{AOC}$ is 90Â°.

Summing up, then:

$\vec{v} \perp \vec{w} \iff \vec{v} \cdot \vec{w} = 0$

which is the property we used in the previous post. What a ride, whew!

I find this fact awesome: by just doing some very simple arithmetics over the coordinates of the vectors we can easily establish if theyâ€™re perpendicular as geometric objects.

Enough for todayâ€¦ stay safe folks!

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