The points $A$, $B$, $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ respectively

Question

The points $A$, $B$, $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ respectively. Given that non-zero numbers $\lambda $, $\mu $ are such that $\lambda \mathbf{a}+\mu \mathbf{b}+\mathbf{c}=0$ and $\lambda +\mu +1=0$. Show that $A$, $B$, $C$ are collinear. Show that the point $P$ with position vector $\mathbf{p}$ given by $\mathbf{p}=4\mathbf{a}-3\mathbf{b}$ lies on $BA$ produced, and find the ratio $PA:PB$. Referring to the origin $O$, two planes ${{\prod }_{1}}$ and ${{\prod }_{2}}$ are given by ${{\prod }_{1}}:\mathbf{r}\cdot \left( \begin{matrix} 1 \ 2 \ -4 \ \end{matrix} ight)=13$ and ${{\prod }_{2}}:\mathbf{r}\cdot \left( \begin{matrix} 1 \ 3 \ 3 \ \end{matrix} ight)=-8$. Given that a point $A\left( 1,7,-10 ight)$ lies on ${{\prod }_{2}}$, show that the perpendicular distance from $A$ to ${{\prod }_{1}}$ is $2\sqrt{21}$. Hence or otherwise find $\overrightarrow{OB}$ where $B$ is the image of $A$ when reflected in the plane ${{\prod }_{1}}$. Write down the Cartesian equations of both ${{\prod }_{1}}$ and ${{\prod }_{2}}$. Find a vector equation of the line of intersection of ${{\prod }_{1}}$ and ${{\prod }_{2}}$. Find a vector equation of the plane which is the image of ${{\prod }_{2}}$ when ${{\prod }_{2}}$ is reflected in ${{\prod }_{1}}$.

Accepted Answer

(a)(i) $A$, $B$, and $C$ are collinear (ii) $PA:PB=3:4$ (b)(i) $2\sqrt{21}$ units (ii) $\overrightarrow{OB}=\begin{pmatrix}-3\-1\6\end{pmatrix}$ (iii) $\Pi_1:x+2y-4z=13$, $\Pi_2:x+3y+3z=-8$ (iv) $\mathbf r=\begin{pmatrix}55\-21\0\end{pmatrix}+t\begin{pmatrix}18\-7\1\end{pmatrix}$ (v) $\mathbf r\cdot\begin{pmatrix}31\83\23\end{pmatrix}=-38$

2008 HCI P1 Q12

2008 HCI P1 Q12

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