2022TrialShort answer

Q15 (b) (ii)

3 marks

Question

$ABCD$ is a parallelogram. $N$ is on $AD$ so that $\overrightarrow{AN}=\frac13\overrightarrow{AD}$ . $M$ is on the diagonal $AC$ . By letting $\overrightarrow{AM}=\lambda\overrightarrow{AC}$ and $\overrightarrow{MB}=\delta\overrightarrow{NB}$ , prove that
$\overrightarrow{AM}=\frac14\overrightarrow{AC}.$

Try the question before revealing the answer.

Solution

(ii) $\overrightarrow{AM}=\lambda\overrightarrow{AC}$

$=\lambda(\vec p+\vec q)$

$\overrightarrow{MB}=\delta\overrightarrow{NB}$

$=\delta\left(\vec p-\frac{1}{3}\vec q\right)$

$\overrightarrow{AB}=\overrightarrow{AM}+\overrightarrow{MB}$
$\vec p+\vec q=\lambda\vec p+\lambda\vec q+\delta\vec p-\frac{1}{3}\delta\vec q$
$\vec p+\vec q=(\lambda+\delta)\vec p+\left(\lambda-\frac{1}{3}\delta\right)\vec q$

Equating $\vec p$ :
$\lambda+\delta=1\quad (1)$

Equating $\vec q$ :
$\lambda-\frac{1}{3}\delta=0\quad (2)$

$\lambda=\frac{1}{3}\delta$
$\frac{4}{3}\delta=1$
$\delta=\frac{3}{4}$

As $\lambda+\delta=1$ , $\lambda=\frac{1}{4}$ .

$\therefore\ \overrightarrow{AM}=\frac{1}{4}\overrightarrow{AC}$

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Syllabus alignment

Vectors

Introduction to three-dimensional vectorsRepresent and operate with vectors in three dimensions using unit, component, ordered-triple and column-vector forms.

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