With reference to the origin $O$, the points $A,B$ and $C$ are such that $\overrightarrow{OA}=\mathb

Question

With reference to the origin $O$, the points $A,B$ and $C$ are such that $\overrightarrow{OA}=\mathbf{a}$, $\overrightarrow{OB}=\mathbf{b}$ and $\overrightarrow{OC}=\mathbf{c}$. The length of $OA$ and $OB$ are $1$ and $2$ respectively. Give the geometrical meaning of $\left| \mathbf{a}	imes \mathbf{b} ight|$. [1] Given that $\mathbf{c}$ is perpendicular to $\mathbf{a}	imes \mathbf{b}$, explain why $\mathbf{c}$ can be expressed as $\mathbf{c}=\mu \mathbf{a}+\lambda \mathbf{b}$ for constants $\mu $ and $\lambda $. [1] Given that angle $AOC=\frac{\pi }{2}$ radians and angle $AOB=\frac{\pi }{3}$ radians and the length of $OC$ is $\sqrt{3}$, show that $\mu =-\lambda $ and hence find all possible value(s) of $\lambda $. [1] Planes ${{p}_{1}}$ and ${{p}_{2}}$ have vector equations $\mathbf{r}\cdot \left( \begin{matrix} \begin{matrix} \sqrt{11} \ 2 \ -1 \\end{matrix} \\end{matrix} ight)=4$ and $\mathbf{r}\cdot \left( \begin{matrix} \begin{matrix} \sqrt{11} \ 2 \ -1 \\end{matrix} \\end{matrix} ight)=12$ respectively. The point $A$ has coordinates $\left( 2\sqrt{11},4,10 ight)$. Find the shortest distance from $A$ to ${{p}_{1}}$. [2] By finding the distance between ${{p}_{1}}$ and ${{p}_{2}}$, or otherwise, determine with clear justification whether $A$ is in between the planes ${{p}_{1}}$ and ${{p}_{2}}$. [2] Given that the point $B$ lies on ${{p}_{1}}$ and has coordinates $(0,-4,-12)$, find the position vector of the point $C$ on ${{p}_{2}}$ such that the line $BC$ is perpendicular to both ${{p}_{1}}$ and ${{p}_{2}}$. [3]

Accepted Answer

(a)(i) Area of parallelogram \text{ or } shortest distance, (a)(ii) $\mathbf{c} = \mu\mathbf{a} + \lambda\mathbf{b}$, (a)(iii) $\lambda = \pm 1$, (b)(i) $4$ units, (b)(ii) A cannot lie between $p_1$ and $p_2$, (b)(iii) $\overrightarrow{OC} = \begin{pmatrix} \frac{\sqrt{11}}{2} \\ -3 \\ -\frac{25}{2} \end{pmatrix}$

2023 TJC P2 Q4

2023 TJC P2 Q4

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