A right opaque with square base $ABCD$ and vertex $V$ is placed at ground level for a shadow dis

Question

A right opaque with square base $ABCD$ and vertex $V$ is placed at ground level for a shadow display, as shown in the diagram. $O$ is the centre of the square base $ABCD$, and perpendicular units vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are in the directions of $\overrightarrow{AB}$, $\overrightarrow{AD}$ and $\overrightarrow{OV}$ respectively. The length of $AB$ is $8$ units and the length of $OV$ is $2h$ units. A point light source for this shadow display is placed at the point $P\left( 20,-4,0 \right)$ and a screen of height $35$ units is placed with its base on the ground such that the screen lies on a plane with vector equation $\mathbf{r}\cdot \left( \begin{matrix} 1 \ 0 \ 0 \ \end{matrix} \right)=\alpha $ where $\alpha <-4$ (see diagram). Find a vector equation of the line depicting the path of the light ray from $P$ to $V$ in terms of $h$.[2] Find an inequality between $\alpha$ and $h$ so that the shadow of the pyramid cast on the screen will not exceed the height of the screen.[3] The point light source is now replaced by a parallel light source whose light rays are perpendicular to the screen and it is also given that $h=10$. Find the exact length of the shadow cast by the edge $VB$ on the screen.[3] A mirror is placed on the plane $VBC$ to create a special effect during the display. Find a vector equation of the plane $VBC$ and hence find the angle of inclination made by the mirror with the ground.[4]

Accepted Answer

(i) $\mathbf{r} = \begin{pmatrix} 20 \ -4 \ 0 \end{pmatrix} + \lambda \begin{pmatrix} -10 \ 2 \ h \end{pmatrix}, \lambda \in \mathbb{R}$ (ii) $h \leq \frac{350}{20-\alpha}$ (iii) $4\sqrt{26}$ (iv) $\mathbf{r} \cdot \begin{pmatrix} 5 \ 0 \ 1 \end{pmatrix} = 20, 78.7^\circ$

2017 RI P1 Q11

2017 RI P1 Q11

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