2022TrialShort answer

Q11 (d) (iii)

3 marks

Question

Using the points defined in part (d), what is the size of the acute angle between $\overrightarrow{AB}$ and $\overrightarrow{CB}$ ? Answer correct to the nearest degree.

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Solution

\overrightarrow{AB}=\begin{bmatrix}2\\1\\-2\end{bmatrix}

\overrightarrow{BC}=\begin{bmatrix}10-9\\14-9\\-4-(-6)\end{bmatrix}=\begin{bmatrix}1\\5\\-10\end{bmatrix}

\cos\theta=\frac{\overrightarrow{AB}\cdot\overrightarrow{CB}}{|\overrightarrow{AB}|\,|\overrightarrow{CB}|}

=\frac{\begin{bmatrix}2\\1\\-2\end{bmatrix}\cdot\begin{bmatrix}1\\5\\10\end{bmatrix}}{\sqrt{2^2+1^2+(-2)^2}\times3\sqrt{14}}

=\frac{2^2+1^2+(-2)\times3\sqrt{14}}{\sqrt{2+5+20}\times3\sqrt{14}} =\frac{3}{\sqrt{14}}

\theta=\cos^{-1}\left(\frac{3}{\sqrt{14}}\right)

\theta\approx37^\circ

Answer correct to the nearest degree.

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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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