Given that $10{\overrightarrow{OF}}\,=3\overrightarrow{OE}+7\overrightarrow{OG}$, show that t

Question

Given that $10{\overrightarrow{OF}}\,=3\overrightarrow{OE}+7\overrightarrow{OG}$, show that the points $E$, $F$ and $G$ are collinear.[2] Relative to the origin $O$ , the points $A$, $B$ and $C$ have position vectors given by $\mathbf{a}=\mathrm{12}\mathbf{i}+\mathrm{2}\mathbf{j}+\mathrm{6}\mathbf{k},$$\mathbf{b}~=\mathrm{5}\mathbf{i}~+\mathbf{j}-\mathrm{2}\mathbf{k}$ and $\mathbf{c}=\left( 1+2\lambda \right)\mathbf{i}+\left( 5+2\lambda \right)\mathbf{j}+\left( -2+\lambda \right)\mathbf{k}$ respectively. The point $P$ lies on $AB$ produced such that $AP=2AB$. Find $\overrightarrow{OP}$[2] If $AP$ is perpendicular to $BC$, find the value of $\lambda$.[3] Using the value of $\lambda $ found in part (ii), find the area of triangle $ACP$.[2]

Accepted Answer

$\overset{\xrightarrow{{}}}{\mathop{OP}}\,=-2\mathbf{i}-10\mathbf{k}$, $\lambda =1$, Area of$\Delta ACP=\frac{1}{2}\sqrt{18696}=68.4$ unit$^{2}$

2013 DHS P1 Q6

2013 DHS P1 Q6

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