Question

Using vectors, find the area of the triangle with vertice A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5) .

Answer 1

 The vertices of the triangle are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
Position vector of A = \[\hat{ i }  + \hat{ j }  + 2 \hat{ k } \]

Position vector of B = \[2 \hat{ i }  + 3 \hat{ j }  + 5 \hat{ k } \]

\[\vec{AB} = \left( 2 \hat{ i } + 3 \hat{ j }  + 5 \hat{ k }  \right) - \left( \hat{ i }  + \hat{ j }  + 2 \hat{ k }  \right) = \hat{ i } + 2 \hat{ j } + 3 \hat{ k } \]

\[\vec{AC} = \left( \hat{ i } 000+ 5 \hat{ j }  + 5 \hat{ k }  \right) - \left( \hat{ i }  + \hat{ j } + 2 \hat{ k }  \right) = 4 \hat{ j }  + 3 \hat{ k } \]

Now, 

\[\vec{AB} \times \vec{AC} = \begin{vmatrix}\hat{ i } & \hat{ j }& \hat{ k } \\ 1 & 2 & 3 \\ 0 & 4 & 3\end{vmatrix} = - 6 \hat{ i }  - 3 \hat{ j } + 4 \hat{ k } \]

∴ Area of ∆ABC =\[\frac{1}{2}\left| \vec{AB} \times \vec{AC} \right|\]

\[= \frac{1}{2}\left| - 6 \hat{ i }  - 3 \hat{ j }  + 4 \hat{ k }  \right|\]

\[ = \frac{1}{2}\sqrt{\left( - 6 \right)^2 + \left( - 3 \right)^2 + 4^2}\]

\[ = \frac{1}{2}\sqrt{36 + 9 + 16}\]

\[ = \frac{\sqrt{61}}{2} \text{ square units } \]