Click here to do these on a digital whiteboard (will not work on small screens).
Q1
Answer
A line $L$ has equation $\mathbf{r} = \begin{pmatrix}2\\-1\\3\end{pmatrix} + t\begin{pmatrix}1\\2\\-2\end{pmatrix}$ and a plane $\Pi$ has equation $2x + y - z = 5$.
Find the point of intersection of $L$ and $\Pi$.
Find the acute angle between $L$ and $\Pi$.
Find the shortest distance from the point $P(2, -1, 3)$ to the plane $\Pi$.
Point of intersection: $(3, 1, 1)$
Acute angle $\approx 11.5°$
Shortest distance $= \dfrac{2}{\sqrt{6}}$
Q2
Answer
The vectors $\mathbf{u} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\mathbf{v} = \mathbf{i} - 2\mathbf{j} + a\mathbf{k}$, where $a$ is a constant.
Find the value of $a$ for which $\mathbf{u}$ and $\mathbf{v}$ are perpendicular.
Given instead that $a = 1$, find $\mathbf{u} \times \mathbf{v}$.
Hence find a unit vector perpendicular to both $\mathbf{u}$ and $\mathbf{v}$.
The point $A$ has position vector $2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ and the line $L$ has equation $\mathbf{r} = \begin{pmatrix}1\\3\\-1\end{pmatrix} + t\begin{pmatrix}2\\-1\\2\end{pmatrix}$.
Find the shortest distance from $A$ to $L$.
Find the reflection of $A$ in $L$.
Shortest distance $= \sqrt{6}$
Reflection of $A$: $(4, 5, -1)$
Q4
Answer
A rhombus $OACB$ is shown in an Argand diagram where $O$ is the origin. The point $A$ has position vector $\mathbf{a} = \begin{pmatrix}4\\1\\0\end{pmatrix}$ and $B$ has position vector $\mathbf{b} = \begin{pmatrix}1\\3\\0\end{pmatrix}$.
Verify that $OACB$ is a rhombus.
Find the angle $OAC$.
Find the area of the rhombus.
$|\mathbf{a}| = |\mathbf{b}| = \sqrt{17}$, confirming all sides are equal