Dot Product of Perpendicular Vectors

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\mathbf a$ and $\mathbf b$ be vector quantities such that $\mathbf a \ne \bszero$ and $\mathbf b \ne \bszero$.

Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ and $\mathbf b$.


Then:

$\mathbf a \cdot \mathbf b = 0$

if and only if:

$\mathbf a$ and $\mathbf b$ are perpendicular.


Proof

By definition of dot product:

$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$

where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$.

When $\mathbf a$ and $\mathbf b$ be perpendicular, by definition $\theta = 90 \degrees$.

The result follows by Cosine of Right Angle, which gives that $\cos 90 \degrees = 0$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Dot_Product_of_Perpendicular_Vectors&oldid=626347"