Feb/Mar 2017 p12 q6
2183
Relative to an origin O, the position vectors of the points A and B are given by
\(\overrightarrow{OA} = 2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}\) and \(\overrightarrow{OB} = 7\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\).
- Use a scalar product to find angle \(OAB\).
- Find the area of triangle \(OAB\).
Solution
(i) First, find the vector \(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = -5\mathbf{i} - \mathbf{j} + 2\mathbf{k}\).
Calculate the dot product \(\overrightarrow{OA} \cdot \overrightarrow{BA} = (2)(-5) + (3)(-1) + (5)(2) = -10 - 3 + 10 = -3\).
Find the magnitudes: \(|\overrightarrow{OA}| = \sqrt{2^2 + 3^2 + 5^2} = \sqrt{38}\) and \(|\overrightarrow{BA}| = \sqrt{(-5)^2 + (-1)^2 + 2^2} = \sqrt{30}\).
Use the cosine formula: \(\cos OAB = \frac{-3}{\sqrt{38} \times \sqrt{30}}\).
Calculate \(OAB = \cos^{-1}\left(\frac{-3}{\sqrt{38} \times \sqrt{30}}\right) \approx 95.1^\circ\) (or 1.66 radians).
(ii) The area of triangle \(OAB\) is given by \(\frac{1}{2} |\overrightarrow{OA}| |\overrightarrow{BA}| \sin OAB\).
Substitute the values: \(\frac{1}{2} \times \sqrt{38} \times \sqrt{30} \times \sin 95.1^\circ \approx 16.8\).
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