For the following vectors, find their magnitude and the angle they make with the positive $x$ axis  - $\mathbf{i} + 4\mathbf{j}$
- $-3\mathbf{i} - 2\mathbf{j}$
     - Magnitude $\sqrt{17}$, angle: $76.0^{\circ}$
- Magnitude $\sqrt{13}$, angle: $-146^{\circ}$
       Given that $(3a+b)\mathbf{i} + \mathbf{j} + ac\mathbf{k} = 7\mathbf{i} - b\mathbf{j} + 4\mathbf{k}$, find the values of $a$, $b$ and $c$.     
  $a = \dfrac{8}{3}$, $b = -1$, $c = \dfrac{3}{2}$   
      The points $A$, $B$ and $C$ have position vectors $\begin{pmatrix}1\\0\\0\end{pmatrix}$, $\begin{pmatrix}5\\3\\4\end{pmatrix}$, and $\begin{pmatrix}2\\-1\\8\end{pmatrix}$.  - Determine if triangle $ABC$ is isosceles.
- Find the area of triangle $ABC$.
     - $|AB| = |BC|$ only, isosceles
- $\dfrac{7}{2}\sqrt{33}$
       Given that $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ and $\mathbf{b} = \mathbf{i} - 2\mathbf{j}$, find:  - $\lambda$ such that $\mathbf{a} + \lambda\mathbf{b}$ is parallel to $\mathbf{i} - \mathbf{j}$
- $\mu$ such that $\mathbf{a} + \mu\mathbf{b}$ is parallel to $3\mathbf{i} + 4\mathbf{j}$.
        Given that $\mathbf{a} = \begin{pmatrix}p\\-q\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}q\\p\end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix}7\\4\end{pmatrix}$, find $p$ and $q$ if $2\mathbf{a} + \mathbf{b} = \mathbf{c}$.     
  $p = 3.6$ and $q = -0.2$   
      Given $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j}$, $\mathbf{b} = p\mathbf{i} - 2p\mathbf{j}$ and that the resultant is parallel to the vector $5\mathbf{i} - 8\mathbf{j}$. Find the resultant vector.     
  $p = 7$, resultant: $10\mathbf{i} - 16\mathbf{j}$   
      Given that $\mathbf{a} = p\mathbf{i} + q\mathbf{j}$ has $|\mathbf{a}| = 10$ and it makes an angle $\theta$ with the positive $x$ axis such that $\sin\theta = \dfrac{3}{5}$, find the possible values of $p$ and $q$.     
  $p = 8$, $q = 6$   
      The point $A$ lies on the circle $x^2 + y^2 = 9$. Given that $\overrightarrow{OA} = 2k\mathbf{i} + k\mathbf{j}$, find the exact values $k$ could be.     
  $k = \pm\dfrac{3\sqrt{5}}{5}$   
      In the following diagram, $P$, $Q$ and $R$ are the midpoints of the sides shown. Given that $\overrightarrow{OP} = \mathbf{p}$ and $\overrightarrow{OR} = \mathbf{r}$, show that triangles $OAB$ and $PQR$ are similar.   
  
     $\overrightarrow{BQ} = \mathbf{p}-\mathbf{r}$ <br> $\overrightarrow{OB} = 2\mathbf{r}$ and $\overrightarrow{PQ} = \mathbf{r}$ so these are parallel and the angles are all the same   
      In the following diagram of a trapezium, $OB$ is parallel to $AC$, and $BD:DA=1:2$. Given that $\overrightarrow{OA} = 4\mathbf{a}$, $\overrightarrow{OB} = 3\mathbf{b}$ and $\overrightarrow{AC} = 6\mathbf{b}$, show that $ODC$ is a straight line and find the ratio $OD:DC$.   
  
     $\overrightarrow{OC} = 4\mathbf{a} + 6\mathbf{b}$, $\overrightarrow{OD} = \frac{1}{3}\overrightarrow{OC}$. <br> Ratio $1:2$, parallel and share a point.   
      $\overrightarrow{OA}$ is the vector $4\mathbf{i} - \mathbf{j} - 2\mathbf{k}$, and $\overrightarrow{OB}$ is the vector $-2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$. Find the unit vector in the direction of $\overrightarrow{AB}$.     
  $\dfrac{1}{\sqrt{61}}(-6\mathbf{i} + 4\mathbf{j} + 3\mathbf{k})$   
      $A$, $B$, and $C$ have vectors $\begin{pmatrix}8\\-7\\4\end{pmatrix}$, $\begin{pmatrix}8\\-3\\3\end{pmatrix}$, and $\begin{pmatrix}12\\-6\\3\end{pmatrix}$  - Determine if triangle $ABC$ is isosceles.
- Find the area of triangle $ABC$.
     - $|AB| = \sqrt{17}$, $|AC| = \sqrt{18}$, $|BC| = 5$. Scalene.
- $8.38$
       $OAB$ is a triangle. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. The point $M$ divides $OA$ in the ratio $2:1$. $MN$ is parallel to $OB$. Show that $AN:NB = 1:2$.   
  
     $\overrightarrow{ON} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$ <br> $\overrightarrow{AN} = \frac{1}{3}(\mathbf{b}-\mathbf{a})$ <br> $\overrightarrow{NB} = \frac{2}{3}(\mathbf{b}-\mathbf{a})$ so ratio is $1:2$   
      $\overrightarrow{PQ} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$, and $\overrightarrow{QR} = -2\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$  - Find angle $PQR$.
- Find the area of triangle $PQR$.
        $OABC$ is a square. $M$ is the midpoint of $OA$ and $Q$ divides $BC$ in the ratio $1:3$. $AC$ and $MQ$ meet at the point $P$. By writing $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$, show that $P$ divides $AC$ in the ratio $2:3$.   
  
     $\overrightarrow{OM} = \frac{1}{2}\mathbf{a}$ <br> $\overrightarrow{MQ} = \frac{1}{4}\mathbf{a} + \mathbf{c}$ <br> $\overrightarrow{AC} = \mathbf{c} - \mathbf{a}$ <br> $\overrightarrow{OP} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{c}$ <br> $\overrightarrow{AP} = \frac{2}{5}(\mathbf{c} - \mathbf{a})$ and $\overrightarrow{PC} = \frac{3}{5}(\mathbf{c} - \mathbf{a})$   
      $\overrightarrow{AB} = 7\mathbf{i} - \mathbf{j} + 2\mathbf{k}$, and $\overrightarrow{BC} = -\mathbf{i} + 5\mathbf{k}$. The point $D$ is such that $\overrightarrow{AD} = 3\overrightarrow{AB}$, and the point $E$ such that $\overrightarrow{AE} = 3\overrightarrow{AC}$.  - Find the area of triangle $ABC$.
- Find the area of triangle $ADE$.
     - $18.7$
- $3^2\times 18.7... = 168$
       Given the points $A$, $B$, $C$ and $D$ have coordinates $(2,-5,-8)$, $(1,-7,-3)$, $(0,15,-10)$ and $(2,19,-20)$ respectively, show that $ABCD$ is a trapezium.     
  $AB = -0.5CD$ so they are parallel. No other sides are parallel, so trapezium.   
      $P$, $Q$ and $R$ have coordinates $(4,-9,-3)$, $(7,-7,-7)$ and $(8,-2,0)$. Find the coordinates of $S$ such that $PQRS$ is a parallelogram.     
  $(5,-4,4)$   
      Given $A$, $B$, $C$ and $D$ have coordinates $(7,12,-1)$, $(11,2,-9)$, $(14,-14,3)$ and $(8,1,15)$ respectively, describe the quadrilateral $ABCD$.     
  $CD = -1.5AB$, 1 pair of parallel sides so trapezium.   
      $OABC$ is a parallelogram and the point $M$ is the midpoint of $AB$. The point $N$ lies on the diagonal $AC$ such that $AN:NC = 1:2$. 
 Given that $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$, show that $OMN$ is a straight line.     
  $\overrightarrow{ON} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{c}$ <br> $\overrightarrow{NM} = \frac{1}{2}\overrightarrow{ON}$. <br> Parallel and share a point, so straight line.   
      The points $A$ and $B$ have position vectors $10\mathbf{i} - 23\mathbf{j} + 10\mathbf{k}$ and $p\mathbf{i} + 14\mathbf{j} - 22\mathbf{k}$ respectively, relative to a fixed origin $O$. Given $OAB$ is an isosceles triangle, find three possible values for $p$.     
  $p = \pm7, \dfrac{1813}{20}$   
      The following diagram shows a trapezium where $AD$ is parallel to $BC$.  - Find $k$.
- Find and simplify an expression for $\overrightarrow{AB}$.
  
       - $-6$
- $-\mathbf{a} - 7\mathbf{b}$
       In the following diagram, $OB:BE = 1:2$, $OC:CA = 1:2$ and $BD:DA = 1:3$. Given that $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$,  - Show that $CDE$ is a straight line, and find the ratio $CD:DE$.
- Show that $BC$ is parallel to $EA$, and find the ratio $BC:EA$.
  
       - $\overrightarrow{CD} = -\frac{1}{12}\mathbf{a} + \frac{3}{4}\mathbf{b}$, $\overrightarrow{DE} = 3\overrightarrow{CD}$ ratio: $1:3$
- $\overrightarrow{BC} = \frac{1}{3}\mathbf{a} - \mathbf{b}$, $\overrightarrow{EA} = 3\overrightarrow{BC}$ ratio: $1:3$
       Given that $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$, $\overrightarrow{OC} = \mathbf{2a}$ and $\overrightarrow{OD} = 2\mathbf{a} + \mathbf{b}$, show that $E$ lies on the line $AB$ if $\overrightarrow{OE} = \frac{1}{3}\overrightarrow{OD}$.     
  $\overrightarrow{AE} = \frac{1}{3}(\mathbf{b} - \mathbf{a})$ and $\overrightarrow{EB} = 2\overrightarrow{AE}$ so they are parallel and share a point.