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Question
Using vectors find the area of the triangle with vertices, A (2, 3, 5), B (3, 5, 8) and C (2, 7, 8).
Solution
\[\text { Let } \vec{a} , \vec{b} \text{ and } \vec{c}^{} \text{ be the position vectors of A, B and C, respectively . Then, } \]
\[ \vec{a} = 2 \hat{ i } + 3 \hat{ j } + 5 \hat{ k } \]
\[ \vec{b} = 3 \hat{ i } + 5\hat{ j } + 8 \hat{ k } \]
\[ \vec{c} = 2 \hat{ i } + 7 \hat{ j } + 8 \hat{ k } \]
\[\text{ Now } , \]
\[ \vec{AB} = \vec{b} - \vec{a} \]
\[ = \hat{ i } + 2 \hat{ j } + 3 \hat{ k } \]
\[ \vec{AC} = \vec{c} - \vec{a} \]
\[ = 0 \hat{ i }+ 4 \hat{ j } + 3 \hat{ k } \]
\[ \therefore \vec{AB} \times \vec{AC} = \begin{vmatrix}\hat{ i }& \hat{ j } & \hat{ k } \\ 1 & 2 & 3 \\ 0 & 4 & 3\end{vmatrix}\]
\[ = - 6 \hat{ i } - 3 \hat{ j } + 4 \hat{ k } \]
\[ \Rightarrow \left| \vec{AB} \times \vec{AC} \right| = \sqrt{36 + 9 + 16}\]
\[ = \sqrt{61}\]
\[\text{ Area of triangleABC } =\frac{1}{2}\left| \vec{AB} \times \vec{AC} \right|\]
\[ =\frac{\sqrt{61}}{2} \text{ sq. units } \]
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