Relative to an origin O, the position vectors of points A and B are given by
\(\overrightarrow{OA} = \begin{pmatrix} 3 \\ -6 \\ p \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 2 \\ -6 \\ -7 \end{pmatrix}\),
and angle \(AOB = 90^\circ\).
(i) Find the value of \(p\).
The point C is such that \(\overrightarrow{OC} = \frac{2}{3} \overrightarrow{OA}\).
(ii) Find the unit vector in the direction of \(\overrightarrow{BC}\).
Solution
(i) Since \(\angle AOB = 90^\circ\), the dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\).
\(\overrightarrow{OA} = \begin{pmatrix} 3 \\ -6 \\ p \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 2 \\ -6 \\ -7 \end{pmatrix}\).
Calculate the dot product:
\(3 \times 2 + (-6) \times (-6) + p \times (-7) = 0\)
\(6 + 36 - 7p = 0\)
\(42 = 7p\)
\(p = 6\)
(ii) \(\overrightarrow{OC} = \frac{2}{3} \overrightarrow{OA} = \frac{2}{3} \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\)
\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix} - \begin{pmatrix} 2 \\ -6 \\ -7 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 11 \end{pmatrix}\)
Magnitude of \(\overrightarrow{BC} = \sqrt{0^2 + 2^2 + 11^2} = \sqrt{125}\)
Unit vector in the direction of \(\overrightarrow{BC} = \frac{1}{\sqrt{125}} \begin{pmatrix} 0 \\ 2 \\ 11 \end{pmatrix}\)
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