Relative to an origin O, the position vectors of points P and Q are given by
\(\overrightarrow{OP} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\) and \(\overrightarrow{OQ} = \begin{pmatrix} 2 \\ 1 \\ q \end{pmatrix}\),
where \(q\) is a constant.
- In the case where \(q = 3\), use a scalar product to show that \(\cos POQ = \frac{1}{7}\).
- Find the values of \(q\) for which the length of \(\overrightarrow{PQ}\) is 6 units.
Solution
(i) To find \(\cos POQ\), we use the scalar product formula:
\(\overrightarrow{OP} \cdot \overrightarrow{OQ} = (-2)(2) + (3)(1) + (1)(3) = -4 + 3 + 3 = 2\).
The magnitudes are \(|\overrightarrow{OP}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{14}\) and \(|\overrightarrow{OQ}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}\).
Thus, \(\cos \theta = \frac{\overrightarrow{OP} \cdot \overrightarrow{OQ}}{|\overrightarrow{OP}| |\overrightarrow{OQ}|} = \frac{2}{\sqrt{14} \times \sqrt{14}} = \frac{2}{14} = \frac{1}{7}\).
(ii) The vector \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 2 \\ 1 \\ q \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ q-1 \end{pmatrix}\).
The length of \(\overrightarrow{PQ}\) is given by \(\sqrt{4^2 + (-2)^2 + (q-1)^2} = 6\).
Squaring both sides, we have \(16 + 4 + (q-1)^2 = 36\).
Thus, \((q-1)^2 = 16\), giving \(q-1 = 4\) or \(q-1 = -4\).
Therefore, \(q = 5\) or \(q = -3\).
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