Given that $\mathbf{u}$ and $\mathbf{v}$ are non-zero vectors such that $\mathbf{u} imes \mathbf{v}

Question

Given that $\mathbf{u}$ and $\mathbf{v}$ are non-zero vectors such that $\mathbf{u}	imes \mathbf{v}=\left( \mathbf{u}-\mathbf{v} ight)	imes \left( \mathbf{u}+\mathbf{v} ight)$, what can be deduced about $\mathbf{u}$ and $\mathbf{v}$? [2] Referred to an origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, where $\mathbf{a}$ and $\mathbf{b}$ are not parallel to each other. Point $C$ is the mid-point of $AB$. Point $D$ lies on $OA$ produced such that $OA:OD=1:p$, where $p>1$. Point $E$ lies on $BD$, between $B$ and $D$, such that $BE:ED=p:2$. Given that points $O$, $C$ and $E$ are collinear, find the value of $p$. [4]

Accepted Answer

(a) The equation becomes $\mathbf{u}	imes \mathbf{v}=2\left( \mathbf{u}	imes \mathbf{v} ight)$, which can only happen if $\mathbf{u}	imes \mathbf{v}=\mathbf{0}$, i.e. ${\mathbf{u}}$ and ${\mathbf{v}}$ are parallel (since the vectors are non-zero). (b) $p=\sqrt{2}$ 	ext{ or } $p=1.41$ (to $3$ s.f.) (since $p>1$)

2018 EJC Promo Q5

2018 EJC Promo Q5

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