Question

If the position vectors of the vertices $A, B$ and $C $ of a $\Delta ABC$ are respectively $4 \hat{i} + 7 \hat{j} + 8 \hat{k} , 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ , then the position vector of the point, where the bisector of $\angle A$ meets BC is :

• A. $\frac{1}{2} \left(4 \hat{i } + 8 \hat{j} + 11\hat{k }\right)$
• B. $\frac{1}{3} \left(6 \hat{i } + 11 \hat{j} + 15 \hat{k }\right)$
• C. $\frac{1}{3} \left(6 \hat{i } + 13 \hat{j} + 18 \hat{k }\right)$
• D. $\frac{1}{4} \left(8 \hat{i } + 14 \hat{j} + 19 \hat{k }\right)$

Answer 1

Answer: (C)

From given data, we plot the graph below

$D$ is the point of intersection of internal angle biscctor of $A$ with line $BC$.
Now, $\overline{A B}=(2 \hat{i}+3 \hat{j}+4 \hat{k})-(4 \hat{i}+7 \dot{j}+8 \hat{k})=-2 \hat{i}-4 \hat{j}-4 \hat{k}$
And $\overrightarrow{A C}=(2 \hat{i}+5 \hat{j}+7 \hat{k})-(4 \hat{i}+7 \hat{j}+8 \hat{k})=-2 \hat{i}-2 \hat{j}-\hat{k}$
$|A \vec{B}|=6,|\overrightarrow{A C}|=3 $
$\frac{B D}{C D}=\frac{A B}{A C}=\frac{6}{3}=\frac{2}{1}$
Therefore, position vector of $D$ Point is
$\frac{(2(2 i+5 j+7 k)+1(2 i+3 j+4 k))}{3}=\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \dot{k})$