9709 P13 - Jun 2012 - Q2
2205
Relative to an origin O, the position vectors of the points A, B and C are given by
\(\overrightarrow{OA} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} \quad \text{and} \quad \overrightarrow{OC} = \begin{pmatrix} 1 \\ 3 \\ p \end{pmatrix}.\)
Find
(i) the unit vector in the direction of \(\overrightarrow{AB}\),
(ii) the value of the constant \(p\) for which angle \(BOC = 90^\circ\).
Solution
(i) To find \(\overrightarrow{AB}\), calculate \(\overrightarrow{OB} - \overrightarrow{OA}\):
\(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}.\)
The modulus of \(\overrightarrow{AB}\) is \(\sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = 7.\)
The unit vector in the direction of \(\overrightarrow{AB}\) is \(\frac{1}{7} \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}.\)
(ii) For \(\angle BOC = 90^\circ\), \(\overrightarrow{OB} \cdot \overrightarrow{OC} = 0.\)
Calculate the dot product:
\(\overrightarrow{OB} \cdot \overrightarrow{OC} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ p \end{pmatrix} = 4 \times 1 + 2 \times 3 + (-2) \times p = 4 + 6 - 2p.\)
Set the dot product to zero:
\(4 + 6 - 2p = 0\)
\(10 = 2p\)
\(p = 5\)
