Express $\overrightarrow{\text{AB}}$ in terms of $m$.

Find the value of $m$.

Consider a unit vector $\mathit{u}$, such that $\mathit{u}=p\mathit{i}-\frac{2}{3}\mathit{j}+\frac{1}{3}\mathit{k}$, where $p>0$.

Point $\text{C}$ is such that $\overrightarrow{\text{BC}}=9\mathit{u}$.

Find the coordinates of $\text{C}$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach to find $\overrightarrow{\text{AB}}$        (M1)

eg     $\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}} , \text{A}-\text{B}$

$\overrightarrow{\text{AB}}=\left(\begin{array}{c}8\\ m-1\\ -8\end{array}\right)$       A1     N2

[2 marks]

valid approach        (M1)

eg     $L=\left(\begin{array}{c}9\\ m\\ -6\end{array}\right) , \left(\begin{array}{c}9\\ m\\ -6\end{array}\right)=\left(\begin{array}{c}-3\\ -19\\ 24\end{array}\right)+s\left(\begin{array}{c}2\\ 4\\ -5\end{array}\right)$

one correct equation        (A1)

eg       $-3+2s=9, -6=24-5s$

correct value for $s$            A1

eg       $s=6$

substituting their $s$ value into their expression/equation to find $m$       (M1)

eg       $-19+6\times 4$

$m=5$       A1     N3

[5 marks]

valid approach        (M1)

eg     $\overrightarrow{\text{BC}}=\left(\begin{array}{c}9p\\ -6\\ 3\end{array}\right), C=9\mathit{u}+B , \overrightarrow{\text{BC}}=\left(\begin{array}{c}x-9\\ y-5\\ z+6\end{array}\right)$

correct working to find $\text{C}$        (A1)

eg     $\overrightarrow{\text{OC}}=\left(\begin{array}{c}9p+9\\ -1\\ -3\end{array}\right), \text{C}=9\left(\begin{array}{c}p\\ -\frac{2}{3}\\ \frac{1}{3}\end{array}\right)+\left(\begin{array}{c}9\\ 5\\ -6\end{array}\right), y=-1$ and $z=-3$

correct approach to find $\left|\mathit{u}\right|$ (seen anywhere)            A1

eg     $p^2+{\left(-\frac{2}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2 , \sqrt{p^2+\frac{4}{9}+\frac{1}{9}}$

recognizing unit vector has magnitude of $1$        (M1)

eg     $\left|\mathit{u}\right|=1 , \sqrt{p^2+{\left(-\frac{2}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2}=1 , p^2+\frac{5}{9}=1$

correct working        (A1)

eg     $p^2=\frac{4}{9} , p=\pm \frac{2}{3}$

$p=\frac{2}{3}$            A1

substituting their value of $p$        (M1)

eg     $\left(\begin{array}{c}x-9\\ y-5\\ z+6\end{array}\right)=\left(\begin{array}{c}6\\ -6\\ 3\end{array}\right), \text{C}=\left(\begin{array}{c}6\\ -6\\ 3\end{array}\right)+\left(\begin{array}{c}9\\ 5\\ -6\end{array}\right), \text{C}=9\left(\begin{array}{c}\frac{2}{3}\\ -\frac{2}{3}\\ \frac{1}{3}\end{array}\right)+\left(\begin{array}{c}9\\ 5\\ -6\end{array}\right), x-9=6$

$\text{C}\left(15, -1, -3\right)$  (accept $\left(\begin{array}{c}15\\ -1\\ -3\end{array}\right)$)     A1     N4

Note: The marks for finding $p$ are independent of the first two marks.
For example, it is possible to award marks such as (M0)(A0)A1(M1)(A1)A1 (M0)A0 or (M0)(A0)A1(M1)(A0)A0 (M1)A0.

[8 marks]