If aperpb then a·b=0 . Does the converse hold?

If $\vec{a}\perp\vec{b}$ then $\vec{a}\cdot\vec{b}=0$. Does the converse hold?

Explanation of Method

Perpendicularity and a zero scalar product — the statement and its converse.

Step 1: Direct statement

If $\vec{a}\perp\vec{b}$, then $\theta = 90^\circ$ and

\[ \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos 90^\circ = 0. \qquad\blacksquare \]

Step 2: The converse

The converse holds for non-zero vectors. If $\vec a\cdot\vec b=0$ with $\vec a\neq\vec 0$, $\vec b\neq\vec 0$, then

\[ \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = 0, \qquad \theta\in[0^\circ,180^\circ] \;\Longrightarrow\; \theta = 90^\circ, \] so $\vec{a}\perp\vec{b}$. $\blacksquare$

Exception: if $\vec a=\vec 0$ (or $\vec b=\vec 0$), then $\vec a\cdot\vec b=0$ for every $\vec b$, although the zero vector has no direction. Many texts adopt the convention that $\vec 0$ is perpendicular to every vector; with it the equivalence holds without exception.

\[\vec{a}\neq\vec{0},\ \vec{b}\neq\vec{0}:\qquad \vec{a}\perp\vec{b} \iff \vec{a}\cdot\vec{b}=0\]

$\vec a$ $\vec b$ $O$

$\theta=90^\circ$: $\vec a\cdot\vec b=0$

$\vec b$ $\vec a=\vec 0$

$\vec a=\vec 0$: $\vec a\cdot\vec b=0$, direction undefined

$\theta$ $\vec a$ $\vec b$ $O$

$\theta\neq90^\circ$: $\vec a\cdot\vec b\neq0$

Step 3: Application: the orths

\[ \vec{i}\cdot\vec{j} = \vec{j}\cdot\vec{k} = \vec{i}\cdot\vec{k} = 0, \qquad \vec{i}\cdot\vec{i} = \vec{j}\cdot\vec{j} = \vec{k}\cdot\vec{k} = 1. \]

Result

Case	$\vec a\cdot\vec b$	Conclusion
$\vec a\perp\vec b$, both non-zero	$0$	$\theta=90^\circ$
$\vec a\cdot\vec b=0$, both non-zero	$0$	$\vec a\perp\vec b$
$\vec a=\vec 0$	$0$	angle undefined (perpendicular by convention)