The points $A, B$ and $C$ have position vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ respectively. (a) Show that the area of triangle $A B C$ is $\frac{1}{2}|\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}|$. Hence show that the shortest distance from $B$ to $A C$ is $$ \frac{|\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}|}{|\mathbf{c}-\mathbf{a}|} $$
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(b) Given that $\mathbf{a}$ and $\mathbf{b}$ are non-zero vectors such that $|\mathbf{a}-\mathbf{b}|=|\mathbf{a}+\mathbf{b}|$, find the value of $\mathbf{a} \cdot \mathbf{b}$.
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(b) 0
(a)$\text { area of triangle } A B C =\frac{1}{2}|(\mathbf{b}-\mathbf{a}) \times(\mathbf{c}-\mathbf{a})| $
$=\frac{1}{2}|(\mathbf{b}-\mathbf{a}) \times \mathbf{c}-(\mathbf{b}-\mathbf{a}) \times \mathbf{a}| $ $=\frac{1}{2}|\mathbf{b} \times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}+\mathbf{a} \times \mathbf{a}| $ $=\frac{1}{2}|\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}+\mathbf{a} \times \mathbf{b}+\mathbf{0}| $ $=\frac{1}{2}|\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}| \quad(\text { shown })$ Let shortest distance from $B$ to $A C$ be $h$ (which is also the perpendicular distance from $B$ to $A C$ ). $A C=|\overrightarrow{A C}|=|\mathbf{c}-\mathbf{a}|$
Let shortest distance from $B$ to $A C$ be $h$ (which is also the perpendicular distance from $B$ to $A C$ ).
$(\mathbf{a}-\mathbf{b}) \cdot(\mathbf{a}-\mathbf{b})=(\mathbf{a}+\mathbf{b}) \cdot(\mathbf{a}+\mathbf{b}) $ $|\mathbf{a}|^2-2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2=|\mathbf{a}|^2+2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2 $ $\mathbf{a} \cdot \mathbf{b}=0$ Alternative Consider a parallelogram $O A C B$ with $\overrightarrow{O A}=\mathbf{a}$ and $\overrightarrow{O B}=\mathbf{b}$. Then the lengths of its diagonals are given by $A B=|\mathbf{a}-\mathbf{b}|$ and $O C=|\mathbf{a}+\mathbf{b}|$. If $|\mathbf{a}-\mathbf{b}|=|\mathbf{a}+\mathbf{b}|$, then $O A C B$ forms a rectangle and thus $\mathbf{a} \cdot \mathbf{b}=0$.
