## Calculus 8th Edition

$|a \cdot b|\leq |a| |b|$
Theorem $3$ states that: $a \cdot b=|a||b| cos \theta$ .... (1) As $|a| \gt 0$ and $|b| \gt 0$ , then $|a \cdot b|=||a||b| cos \theta|=|a||b| |cos \theta|$ .... (2) (i) If $0 \leq \theta \leq \pi/2$, then If $0 \leq |a||b| |cos \theta| \leq |a||b| |$ (ii) If $\pi/2 \leq \theta \leq \pi$, then If $|a||b| \geq |a \cdot b| | \gt 0$ From the above two conditions, we have found that $|a \cdot b|\leq |a| |b|$