Referred to the origin, points $A$ and $B$ have non-zero position vectors $\mathbf{a}$ and $\mathbf{

Question

Referred to the origin, points $A$ and $B$ have non-zero position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. Show that $\left( \mathbf{a}+\mathbf{b} ight)	imes \left( \mathbf{a}-\mathbf{b} ight)=-2\left( \mathbf{a}	imes \mathbf{b} ight)$. [2] It is given that $\left| \mathbf{b} ight|=2\left| \mathbf{a} ight|$. Find the maximum value of $\left| \left( \mathbf{a}+\mathbf{b} ight)	imes \left( \mathbf{a}-\mathbf{b} ight) ight|$ in terms of $\left| \mathbf{a} ight|$. [3] It is further given that $A$ is a fixed point on the $x$-axis and $B$ is a variable point on the $xz$-plane. Given that the value of $\left| \left( \mathbf{a}+\mathbf{b} ight)	imes \left( \mathbf{a}-\mathbf{b} ight) ight|$ is maximum, write down all possible expressions for $\left( \mathbf{a}+\mathbf{b} ight)	imes \left( \mathbf{a}-\mathbf{b} ight)$ in terms of $\left| \mathbf{a} ight|$. [2]

Accepted Answer

(i) Shown; (ii) 4|a|²; (iii) (0, 4|a|², 0) or (0, -4|a|², 0)

2020 NJC Promo Q7

2020 NJC Promo Q7

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