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If $\overrightarrow {OA}=\hat i+3\hat j-2\hat k\:\:and\:\:\overrightarrow {OB}=3\hat i+\hat j-2\hat k$ and $C$ is a point of $AB$ so that $\overrightarrow {OC}$ bisects the angle $AOB$, then $\overrightarrow {OC}=?$

$\begin{array}{1 1}(a)\:4\hat i+4\hat j-4\hat k\:\:\qquad\:(b)\:2\hat i+2\hat j-2\hat k\:\:\qquad\:(c)\:\hat i+\hat j-\hat k\:\:\qquad\:(d)\:None\:of\:these \end{array} $

1 Answer

Given $\overrightarrow {OA}=\hat i+3\hat j-2\hat k\:\;and\:\:\overrightarrow {OB}=3\hat i+\hat j-2\hat j$
$\Rightarrow\:|\overrightarrow {OA}|=\sqrt {1+9+4}=\sqrt {14}$ and $ |\overrightarrow {OB}|=\sqrt {14}$
$\therefore$ $\overrightarrow {OC}$ divides $AB$ is the ratio $\sqrt {14}:\sqrt {14}=1:1$
$\therefore $ From section formula $\overrightarrow {OC}=\large\frac{\overrightarrow {OA}+\overrightarrow {OB}}{2}=2\hat i+2\hat j-2\hat k$
answered Jan 5, 2014 by rvidyagovindarajan_1

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