Hello. This post will talk about orthogonal (perpendicular) vectors in the n-th dimension \(\mathbb{R}^{n}\). It is assumed that one knows about dot products.
Recall that we can find the cosine of an angle \(\theta\) using the dot product of vectors \(\boldsymbol{u}\) and \(\boldsymbol{v}\) in \(\mathbb{R}^{n}\) and its norms.
The formula for finding the cosine of an angle \(\theta\) is:
\[\displaystyle \cos(\theta) = \dfrac{\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{v}}{||\boldsymbol{u}|| ||\boldsymbol{v}||}\]
Applying the inverse cosine function or the arccos function (which is the same thing) to both sides of the equations, we can isolate the angle \(\theta\).
\[\displaystyle \theta = \cos^{-1}(\dfrac{\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{v}}{||\boldsymbol{u}|| ||\boldsymbol{v}||})\]
or
\[\displaystyle \theta = \arccos(\dfrac{\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{v}}{||\boldsymbol{u}|| ||\boldsymbol{v}||})\]
where \(\theta\) can take values from 0 to \(\pi\) (inclusive).
From the above, if \(\boldsymbol{u} \cdot \boldsymbol{v} = 0\) then \(\theta\) is a right angle (\(\theta\) = 90 degrees or \(\dfrac{\pi}{2}\) in radians). We can also say that the vectors \(\boldsymbol{u}\) and \(\boldsymbol{v}\) or perpendicular or orthogonal in \(\mathbb{R}^{n}\).
The zero vector in \(\mathbb{R}^{n}\) is orthogonal to every vector in \(\mathbb{R}^{n}\).
Proof
Suppose we have the zero vector \(\boldsymbol{0}\) and another arbitrary vector such as ($ a_{1}, a_{2}, , a_{n}$) in \(\mathbb{R}^{n}\). Taking the dot product of these two vectors gives us:
\[\displaystyle \begin{array} {lcl} \boldsymbol{0} \boldsymbol{\cdot} (a_{1}, a_{2}, \dots , a_{n}) & = & (0, 0, 0, \dots, 0) \boldsymbol{\cdot} (a_{1}, a_{2}, \dots , a_{n}) \\ & = & 0 \cdot a_1 + 0 \cdot a_2 + \dots + 0 \cdot a_n \\ & = & 0 + 0 + \dots + 0 \\ & = & 0 \end{array} \]
We illustrate the concept of orthogonal vectors with a few examples.
Example One
In \(\mathbb{R}^{2}\) the dot product of the vectors \(\boldsymbol{a} = (2, -9)\) and \(\boldsymbol{b} = (8, 1)\) is 7. These two vectors are not orthogonal to one another.
\[\displaystyle \begin{array} {lcl} \boldsymbol{a} \boldsymbol{\cdot} \boldsymbol{b} & = & (2, -9) \boldsymbol{\cdot} (8, 1) \\ & = & 2 \times 8 + (- 9) \times 1 \\ & = & 16 - 9\\ & = & 7\\ \end{array} \]
Example Two
Consider two vectors in \(\boldsymbol{c} = (-19, 1, 0, 2)\) and \(\boldsymbol{d} = (0, 20, 32, -10)\) in \(\mathbb{R}^{4}\). Are these two vectors orthogonal (perpendicular) in \(\mathbb{R}^{4}\)?
Answer
To determine whether the two vectors are orthogonal (perpendicular) in \(\mathbb{R}^{4}\), we compute the dot product of vectors \(\boldsymbol{c}\) and \(\boldsymbol{d}\).
\[\displaystyle \begin{array} {lcl} \boldsymbol{c} \boldsymbol{\cdot} \boldsymbol{d} & = & (-19, 1, 0, 2) \boldsymbol{\cdot} (0, 20, 32, -10) \\ & = & -19 \times 0 + 1 \times 20 + 0 \times 32 + 2 \times (-10)\\ & = & 0 + 20 + 0 - 20\\ & = & 0\\ \end{array} \]
The vectors \(\boldsymbol{c}\) and \(\boldsymbol{d}\) are orthogonal to one another in \(\mathbb{R}^{4}\).
Reference
Elementary Linear Algebra (Tenth Edition) by Howard Anton.
The featured image is from https://www.math.hmc.edu/calculus/tutorials/vectoranalysis/gif/vector9.gif.
