Question
The adjacent sides $$AB$$ and $$AC$$ of a triangle $$ABC$$ are represented by the vectors $$ - 2\hat i + 3\hat j + 2\hat k$$ and $$ - 4\hat i + 5\hat j + 2\hat k$$ respectively. The area of the triangle $$ABC$$ is :
A.
$$6$$ square units
B.
$$5$$ square units
C.
$$4$$ square units
D.
$$3$$ square units
Answer :
$$3$$ square units
Solution :
Area of $$\Delta ABC\,{\bf{:}}$$
\[\begin{array}{l}
= \frac{1}{2}\left( {\overrightarrow {AB} \times \overrightarrow {AC} } \right)\\
= \frac{1}{2}\left| \begin{array}{l}
\,\,\,\hat i\,\,\,\,\,\hat j\,\,\,\,\hat k\\
- 2\,\,\,\,3\,\,\,\,2\\
- 4\,\,\,\,5\,\,\,\,2
\end{array} \right|
\end{array}\]
$$\eqalign{
& = \frac{1}{2}\left[ {\hat i\left( {6 - 10} \right) - \hat j\left( { - 4 + 8} \right) + \hat k\left( { - 10 + 12} \right)} \right] \cr
& = \frac{1}{2}\left[ { - 4\hat i - 4\hat j + 2\hat k} \right] \cr
& = \frac{1}{2}\sqrt {16 + 16 + 4} \cr
& = \frac{1}{2}\sqrt {36} \cr
& = \frac{1}{2} \times 6 \cr
& = 3{\text{ square units}} \cr} $$