# The vector from origin to the points $A$ and $B$ are $\overrightarrow{a}=2\hat i-3\hat j-2\hat k$ and $\overrightarrow{b}=2\hat i+3\hat j-\hat k,$ respectively,then the areqa of triangle $OAB$ is

$(A)\;340\quad(B)\;\sqrt {25}\quad(C)\;\sqrt {229}\quad(D)\;\frac{1}{2}\sqrt {229}$

Toolbox:
• Area of $\Delta$ is = $\large\frac{1}{2}$$|\overrightarrow a \times\overrightarrow b | Let \overrightarrow a=2\hat i-3\hat j-2\hat k and \overrightarrow b=2\hat i+3\hat j-\hat k Area of \Delta is = \large\frac{1}{2}$$|\overrightarrow {OA} \times\overrightarrow {OB} |$
$\qquad \qquad =>\large\frac{1}{2}$$|\overrightarrow a \times\overrightarrow b | Let us first determine \overrightarrow a \times\overrightarrow b \overrightarrow a\times\overrightarrow b=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\2&-3&-2\\2&3&-1\end{array}\right| On expanding we get, =(3+6)\hat i-(-2+4)\hat j+(6+6)\hat k On simplifying we get, =9\hat i-2\hat j+12\hat k |\overrightarrow a\times\overrightarrow b|=\sqrt{9^2+(-2)^2+12^2} \qquad\qquad=\sqrt {81+4+144} \qquad\qquad=\sqrt {229} We know that area of the triangle is= \large\frac{1}{2}$$\sqrt {229}$
Area of $\Delta$ OAB=$\frac{1}{2}\sqrt{229}$
Hence the correct option is $D$