The diagram shows a parallelogram \(OABC\) in which
\(\overrightarrow{OA} = \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix}\).
(i) Use a scalar product to find angle \(BOC\).
(ii) Find a vector which has magnitude 35 and is parallel to the vector \(\overrightarrow{OC}\).
Solution
(i) First, find \(\overrightarrow{OC}\) using \(\overrightarrow{OC} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).
Calculate the dot product \(\overrightarrow{OC} \cdot \overrightarrow{OB} = 2 \times 5 + (-3) \times 0 + 6 \times 2 = 22\).
Find the magnitudes: \(|\overrightarrow{OC}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7\) and \(|\overrightarrow{OB}| = \sqrt{5^2 + 0^2 + 2^2} = \sqrt{29}\).
Use the cosine formula: \(\overrightarrow{OC} \cdot \overrightarrow{OB} = |\overrightarrow{OC}| \cdot |\overrightarrow{OB}| \cdot \cos \angle BOC\).
\(22 = 7 \times \sqrt{29} \times \cos \angle BOC\).
\(\cos \angle BOC = \frac{22}{7 \times \sqrt{29}}\).
\(\angle BOC = \cos^{-1}\left(\frac{22}{7 \times \sqrt{29}}\right) \approx 54.3^\circ\) (or 0.948 rad).
(ii) The magnitude of \(\overrightarrow{OC}\) is 7. To find a vector with magnitude 35, scale \(\overrightarrow{OC}\) by \(\frac{35}{7} = 5\).
Thus, the vector is \(\pm 5 \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).
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