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Question
Using vectors, find the area of the triangle with vertice A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1) .
Solution
The vertices of the triangle are A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1).
Position vector of A = \[\hat{ i }+ 2 \hat{ j } + 3 \hat{ k } \]
Position vector of B = \[2 \hat{ i } - \hat{ j }+ 4 \hat{ k } \]
Position vector of C = \[4 \hat{ i } + 5 \hat{ j } - \hat{ k } \]
\[\vec{AB} = \left( 2 \hat{ i } - \hat{ j } + 4 \hat{ k } \right) - \left( \hat{ i } + 2 \hat{ j } + 3 \hat{ k } \right) = \hat{ i } - 3 \hat{ j } + \hat{ k } \]
\[\vec{AC} = \left( 4 \hat{ i } + 5 \hat{ j } - \hat{ k } \right) - \left( \hat{ i } + 2 \hat{ j } + 3 \hat{ k } \right) = 3 \hat{ i } + 3 \hat{ j } - 4 \hat{ k } \]
Now,
\[\vec{AB} \times \vec{AC} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & - 3 & 1 \\ 3 & 3 & - 4\end{vmatrix} = 9 \hat{ i } + 7 \hat{ j } + 12 \hat{ k } \]
∴ Area of ∆ABC = \[\frac{1}{2}\left| \vec{AB} \times \vec{AC} \right|\]
\[= \frac{1}{2}\left| 9 \hat{ i } + 7 \hat{ j } + 12 \hat{ k } \right|\]
\[ = \frac{1}{2}\sqrt{9^2 + 7^2 + {12}^2}\]
\[ = \frac{1}{2}\sqrt{81 + 49 + 144}\]
\[ = \frac{\sqrt{274}}{2} \text{ square units } \]
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