2023TrialShort answer

Q11 (e) (ii)

2 marks

Question

Using the diagram and information from part (e), calculate the angle between $\overrightarrow{MD}$ and $\overrightarrow{ON}$ .

Try the question before revealing the answer.

Solution

Let the required angle be $\theta$ .

\overrightarrow{MD}\cdot\overrightarrow{ON}=|MD||ON|\cos\theta

\cos\theta=\frac{\overrightarrow{MD}\cdot\overrightarrow{ON}}{|\overrightarrow{MD}|\,|\overrightarrow{ON}|}

=\frac{0-6+3}{\sqrt{17}\sqrt{10}}

=\frac{3}{\sqrt{170}}

\Rightarrow\theta=76^\circ42'\text{ to nearest minute}

$76.7^\circ$ , $103^\circ18'$ ; $(103.3^\circ$ also accepted as question was not very discriminating).

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Syllabus alignment

Vectors

Further operations with three-dimensional vectorsUse vector magnitude, unit vectors and scalar products to solve problems and prove geometric results.

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