Relative to the origin $O$, two points $A$ and $B$ have position vectors given by $\mathbf{a}$ and $

Question

Relative to the origin $O$, two points $A$ and $B$ have position vectors given by $\mathbf{a}$ and $\mathbf{b}$ respectively such that they are non-zero, non-parallel and $\left| \mathbf{a} ight|=\left| \mathbf{b} ight|$. Show that $\left( \mathbf{a}+\mathbf{b} ight)\cdot \left( \mathbf{a}-\mathbf{b} ight)=0$. Give a geometrical interpretation of the result shown in part (ii). Suppose $\left| \mathbf{a} ight|=\left| \mathbf{b} ight|=1$, show that ${{\left| \mathbf{a}\cdot \mathbf{b} ight|}^{2}}+{{\left| \mathbf{a}	imes \mathbf{b} ight|}^{2}}=1$.

Accepted Answer

(i) Proven, (ii) OACB is a rhombus, (iii) Proven

2016 NJC CT Q6

2016 NJC CT Q6

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