If 4$\widehat{\dot{\mathbf{i}}} + 7\widehat{\dot{\mathbf{j}}} + 8\widehat{\mathbf{k}},\ 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}}$ and $2\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + 7\widehat{\mathbf{k}}$ are the position vectors of the vertices $A,B\ and\ C$ respectively of triangle $ABC$. The position vector of the point where the bisector of angle $A$ meets $BC$ is

Question

Dr. Abubacker · Accepted Answer

(a)
$\overrightarrow{\mathbf{AB}} = 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}} - 4\widehat{\dot{\mathbf{i}}} - 7\widehat{\dot{\mathbf{j}}} - 8\widehat{\mathbf{k}} = - 2\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}} - 4\widehat{\mathbf{k}}$
and $\overrightarrow{\mathbf{AC}} = 2\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + 7\widehat{\mathbf{k}} - 4\widehat{\dot{\mathbf{i}}} - 7\widehat{\dot{\mathbf{j}}} - 8\widehat{\mathbf{k}} = - 2\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$
$	herefore\left| \overrightarrow{\mathbf{AB}} ight| = 6$ and |$\overrightarrow{\mathbf{AC}}$|=3
$	herefore\$Position vector of required bisector
$= \frac{6\left( 2\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + 7\widehat{\mathbf{k}} ight) + 3\left( 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}} ight)}{6 + 3}$
$= \frac{1}{3}(6\widehat{\dot{\mathbf{i}}} + 13\widehat{\dot{\mathbf{j}}} + 18\widehat{\mathbf{k}})$

Explanation

Reviewed by Dr. Abubacker

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