In the diagram, $O$ is centre of the square base $ABCD$ of a right pyramid, vertex $V$. Perpendicula

Question

In the diagram, $O$ is centre of the square base $ABCD$ of a right pyramid, vertex $V$. Perpendicular unit vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are parallel to $AB$, $AD$, $OV$ respectively. The length of $AB$ is 4 units and the length of $OV$ is $2h$ units. $P$, $Q$, $M$ and $N$ are mid-points of $AB$, $BC$, $CV$ and $VA$ respectively. The point $O$ is taken as the origin for position vectors. Show that the equation of the line $PM$ may be expressed as $\mathbf{r}=\left( \begin{matrix} 0 \ -2 \ 0 \ \end{matrix} ight)+t\left( \begin{matrix} 1 \ 3 \ h \ \end{matrix} ight)$, where $t$ is a parameter. [2] Find the equation for the line $QN$. [2] Show that the lines $PM$ and $QN$ intersect, and that the position vector $\overrightarrow{OX}$ of their point of intersection is $\left( \begin{matrix} \frac{1}{2} \ -\frac{1}{2} \ \frac{1}{2}h \ \end{matrix} ight)$. [3] Given that $OX$ is perpendicular to $VB$, find the value of $h$ and calculate the acute angle between $PM$ and $QN$, giving your answer correct to the nearest $0.1{}^\circ $. [4]

Accepted Answer

(i) l_PM shown (ii) r=(2,0,0)+λ(-3,h,-1) (iii) intersect at X(1/2,1/2,-1/2) (iv) h=√2, θ=70.5°

N1994 P1 Q14

N1994 P1 Q14

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