Orthogonal vectors

Two (non-zero) vectors $\vec{a}$ and $\vec{b}$ are called orthogonal (or perpendicular), written

\vec{a} \perp \vec{b}

if the angle between the vectors is a right angle (that is, $\alpha = 90^\circ$ .

Recall that $\cos(\alpha)=0$ if, and only if $\alpha=90^\circ, 270^\circ, ...$ , so if, and only if the angle is a right angle. And because of the angle formula

\cos(\alpha)=\frac{\vec{a}\bullet \vec{b}}{\vert\vec{a}\vert\cdot \vert\vec{b}\vert}

the following important theorem follows:

\boxed{\vec{a} \perp \vec{b}\text{ if, and only if, } \vec{a}\bullet \vec{b}=0}

Example 1

The vectors $\left(\begin{array}{r} 2\\ 3\\ 4 \end{array}\right)$ and $\left(\begin{array}{r} 0\\ -4\\ 3 \end{array}\right)$ are orthogonal, because

\left(\begin{array}{r} 2\\ 3\\ 4 \end{array}\right) \bullet \left(\begin{array}{r} 0\\ -4\\ 3 \end{array}\right)=2\cdot 0+ 3\cdot(-4)+ 4\cdot 3 =0

Exercise 1

Q1

Are the vectors $\left(\begin{array}{r} 1\\ -2\\ 3 \end{array}\right)$ and $\left(\begin{array}{r} 2\\ 1\\ -3 \end{array}\right)$ orthogonal?

Q2

Find at least one non-zero vector that is orthogonal to $\vec{a}=\left(\begin{array}{r} 3\\ 1\\ -2 \end{array}\right)$ . How many such vectors exist and what geometrical object is formed by all these vectors if attached to the origin?

Q3

Consider the points $A(2\vert 0\vert 0)$ and $B(0 \vert 3 \vert 0)$ , and the straight line $g$ that passes through the origin and has direction $\vec{v}=\left(\begin{array}{r} 0\\ 1\\ 1 \end{array}\right)$ . Find all points $P$ on $g$ such that the segments $AP$ and $BP$ form a right angle.

Solution

A1

$\left(\begin{array}{r} 1\\ -2\\ 3 \end{array}\right) \bullet \left(\begin{array}{r} 2\\ 1\\ -3 \end{array}\right)=1\cdot 2+ (-2)\cdot 1+ 3\cdot (-3) \neq 0$ , so not orthogonal.

A2

Find vector $\vec b$ with

\vec a \bullet \vec b=3b_x+b_y-2b_z=0

e.g. $b_x=0, b_y=1$ , thus $b_z=0.5$ or $b_x=1, b_y=1$ , thus $b_z=2$ , and so on. Just assign $b_x$ and $b_z$ some numbers, and then calculate $b_z$ from the equation above. Clearly, there are infinitely many such vectors.

A3

Find point $P(x\vert y\vert z)$ with $P$ on $g$ and $\overrightarrow{AP} \perp \overrightarrow{BP}$ (see figure).

As $P$ on $g$ , there is a scalar $c$ with
$\vec P = \vec O +c\cdot \vec v$ $\left(\begin{array}{r} x\\ y\\ z \end{array}\right) = \left(\begin{array}{r} 0\\ 0\\ 0 \end{array}\right) + c\cdot \left(\begin{array}{r} 0\\ 1\\ 1 \end{array}\right) = \left(\begin{array}{r} 0\\ c\\ c \end{array}\right)$
and thus
$\begin{array}{r} x=0\\y=c\\z=c \end{array}$
From $\overrightarrow{AP} \perp \overrightarrow{BP}$ follows
$\overrightarrow{AP} \bullet \overrightarrow{BP} =0$ $\left(\begin{array}{r} -2\\ c\\ c \end{array}\right)\bullet \left(\begin{array}{r} 0\\ c-3\\ c \end{array}\right) = 0$ $0+c(c-3)+c^2 = 0$ $2c^2-3c=0$
Solving this quadratic equation we get $c_1=0$ and $c_2=1.5$ . It follows $P_1=\underline{(0 \vert 0 \vert 0)}$ and $P_2=\underline{(0\vert 1.5\vert 1.5)}$ .