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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Vector Algebra
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The position vector of the vertices $A,B,C$ of a $\Delta\: ABC$ are $\hat i-\hat j-3\hat k,\:2\hat i+\hat j-2\hat k\:\:and\:\:-5\hat i+2\hat j-6\hat k$ respectively. The length of the bisector $AD$ of angle $A$ where $D$ is on the line $BC$ is ?

$\begin{array}{1 1} \large (a)\:\frac{15}{2}\;\:\:\qquad\:\:(b)\:\:\frac{11}{2}\:\:\:\qquad\:\:(c)\:\:\frac{1}{4}\:\:\:\qquad\:\:(d)\:\:\frac{3}{4} \sqrt {10} \end{array} $

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  • Angular bisector of an angle of a $\Delta $ divides the opp. side in the ratio of the length of corresponding sides.
Given $\overrightarrow {OA}=\hat i-\hat j-3\hat k,\overrightarrow {OB}=2\hat i+\hat j-2\hat k,\:\overrightarrow {OC}=-5\hat i+2\hat j-6\hat k$
$\overrightarrow {AB}=\hat i+2\hat j+\hat k$ and $\overrightarrow {AC}=-6\hat i+3\hat j-3\hat k$
$\therefore$ $|\overrightarrow {AB}|=\sqrt {6}$ and $|\overrightarrow {AC}|=\sqrt {54}=3\sqrt 6$
Let $AD$ be the angular bisector of angle $A$ where $D$ lies on the line $BC$.
$\Rightarrow$ $D$ divides $BC$ in the ratio $\sqrt 6:3\sqrt 6=1:3$
From section formula position vector of $D$ is given by $\large\frac{\overrightarrow {OC}+3\overrightarrow {OB}}{4}$
$=\large\frac{1}{4}$$(\hat i+5\hat j-12\hat k)$
$\Rightarrow\:\overrightarrow {AD}=\large\frac{1}{4}$$(\hat i+5\hat j-12\hat k)-(\hat i-\hat j-3\hat 3k)=\large\frac{3}{4}$$(-\hat i+3\hat j)$
$\therefore$ The required distance $AD =\sqrt {\large\frac{9}{16}+\frac{81}{16}+0}=\large\frac{3}{4}$$\sqrt {10}$
answered Jan 5, 2014 by rvidyagovindarajan_1

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