2022TrialShort answer

Q15 (b) (i)

1 mark

Question

$ABCD$ is a parallelogram. $N$ is on $AD$ so that $\overrightarrow{AN}=\frac13\overrightarrow{AD}$ . $M$ is on the diagonal $AC$ . Letting $\overrightarrow{AB}=\vec p$ and $\overrightarrow{AD}=\vec q$ , express $\overrightarrow{NB}$ in terms of $\vec p$ and $\vec q$ .

Try the question before revealing the answer.

Solution

(i) $\overrightarrow{AN}=\frac{1}{3}\vec q$

$\therefore\ \overrightarrow{NA}=-\frac{1}{3}\vec q$

$\overrightarrow{NB}=\overrightarrow{NA}+\overrightarrow{AB}$
$=-\frac{1}{3}\vec q+\vec p$
$=\vec p-\frac{1}{3}\vec q$

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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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