The vectors from origin to the points A and B are \vec{A} = 3\hat{i} - 6\hat{j} + 2\hat{k} and \(\vec{\mathbf{B}} = 2\hat{i} + \hat{j} - 2\hat{k}\) respectively. The area of the triangle OAB be
\((a) \frac{5}{2} \sqrt{17} \text{ sq.unit} (b) \frac{2}{5} \sqrt{17} \text{ sq.unit} (c) \frac{3}{5} \sqrt{17} \text{ sq.unit} (d) \frac{5}{3} \sqrt{17} \text{ sq.unit}\)
Text Solution
Verified by ExpertsA
Given \vec{0A} = \vec{a} = 3\hat{i} - 6\hat{j} + 2\hat{k} and \vec{OB} = \vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}
\(\therefore (\vec{a} \times \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & 2 \\ 2 & 1 & -2 \end{vmatrix}\)
= (12 - 2)\hat{i} + (4 + 6)\hat{j} + (3 + 12)\hat{k}
\(= 10\hat{i} + 10\hat{j} + 15\hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{10^{2} + 10^{2} + 15^{2}}\)
\(= \sqrt{425}\) \(= 5 \sqrt{7}\)
Area of \(\Delta OAB = \frac{1}{2} \left| \vec{a} \times \vec{b} \right| = \frac{5 \sqrt{17}}{2}\) sq.unit.
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