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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 $\overrightarrow {OC} $ is angular bisector of angle $AOB$ with $C$ being the point on the line $AB$, then $\overrightarrow {OC}=?$

1 Answer

Since $ |\overrightarrow {OA}|=|\overrightarrow {OB}=\sqrt {14}$
$\Delta AOB$ is isosceles triangle.
$\therefore \:\overrightarrow {OC}$ which is angular bisector is also median of $AB$
$\therefore\:\overrightarrow {OC}=\large\frac{(\hat i+3\hat j-2\hat k)+(3\hat i+\hat j-2\hat k)}{2}$
$=2\hat i+2\hat j-2\hat k$
answered Nov 13, 2013 by rvidyagovindarajan_1

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