Relative to the origin $O$, the fixed points $A$ and $B$ are such that $\overrightarrow{OA}=\mathbf{

Question

Relative to the origin $O$, the fixed points $A$ and $B$ are such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. The point $R$ has position vector $\overrightarrow{OR}=\left( \begin{matrix} x \ y \ z \ \end{matrix} ight)$. Interpret geometrically the point $R$, stating what $d$ represents, given that $\overrightarrow{OR}\cdot \frac{\left( \mathbf{a}	imes \mathbf{b} ight)}{\left| \mathbf{a}	imes \mathbf{b} ight|}=d$ where $d$ is a constant scalar. [2] The points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively relative to the origin $O$. $M$ is the mid- point of $AB$. The point $N$ on $OB$ is such that $ON:NB=k:1-k$, where $0 [2] The lines $OM$ and $AN$ intersect at the point $P$. Given that $OP:PM=2k:1-k$, show that $\overrightarrow{OP}=\frac{k}{1+k}\left( \mathbf{a}+\mathbf{b} ight)$. Hence, find the area of triangle $OPN$ in terms of $k$. [4]

Accepted Answer

(a) R lies on plane parallel to a and b with shortest distance d from origin; (b)(i) OP = k/(k+1)(a+b); (b)(ii) |b| = 2√3

2022 SAJC J2 MYE Q5

2022 SAJC J2 MYE Q5

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