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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra

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

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

1 Answer

  • 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 $D$
answered May 28, 2013 by meena.p

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