Relative to an origin O , the position vectors of the points A , B and X are given by
\(\overrightarrow{OA} = \begin{pmatrix} -8 \\ -4 \\ 2 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 10 \\ 2 \\ 11 \end{pmatrix} \quad \text{and} \quad \overrightarrow{OX} = \begin{pmatrix} -2 \\ -2 \\ 5 \end{pmatrix}.\)
(i) Find \(\overrightarrow{AX}\) and show that AXB is a straight line.
The position vector of a point C is given by \(\overrightarrow{OC} = \begin{pmatrix} 1 \\ -8 \\ 3 \end{pmatrix}.\)
(ii) Show that CX is perpendicular to AX .
(iii) Find the area of triangle ABC .
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Solution
(i) To find \(\overrightarrow{AX}\), calculate \(\overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix} -2 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} -8 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix}.\)
To show AXB is a straight line, find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 10 \\ 2 \\ 11 \end{pmatrix} - \begin{pmatrix} -8 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} 18 \\ 6 \\ 9 \end{pmatrix}.\)
Since \(\overrightarrow{AB} = 3 \overrightarrow{AX}\), AXB is a straight line.
(ii) To show CX is perpendicular to AX , find \(\overrightarrow{CX} = \overrightarrow{OX} - \overrightarrow{OC} = \begin{pmatrix} -2 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ -8 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix}.\)
Calculate the dot product \(\overrightarrow{CX} \cdot \overrightarrow{AX} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix} = -18 + 12 + 6 = 0.\)
Since the dot product is zero, CX is perpendicular to AX .
(iii) To find the area of triangle ABC , calculate the magnitudes \(|\overrightarrow{CX}| = \sqrt{(-3)^2 + 6^2 + 2^2} = 7\) and \(|\overrightarrow{AB}| = \sqrt{18^2 + 6^2 + 9^2} = 21.\)
The area is \(\frac{1}{2} \times |\overrightarrow{CX}| \times |\overrightarrow{AB}| = \frac{1}{2} \times 7 \times 21 = 73.5.\)
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