Question

If \(7\hat{j}+10\hat{k},-\hat{i}+6\hat{j}+6\hat{k}\) and \(-4\hat{i}+9\hat{j}+6\hat{k}\) are the position vectors of the vertices A, B and C repectively of \(△ABC\). Then the position vector of the point where the bisector of the angle A meets side BC is

• A. \((2+3\sqrt{2})\hat{i}+(3+3\sqrt{3})\hat{j}+6\hat{k}\)
• B. \((2-3\sqrt{2})\hat{i}+(3+3\sqrt{2})\hat{j}+6\hat{k}\)
• C. \((2-3\sqrt{2})\hat{i}+(3-3\sqrt{2})\hat{j}+6\hat{k}\)
• D. \((2-3\sqrt{2})\hat{i}+(3+3\sqrt{2})\hat{j}-6\hat{k}\)

Hint:

The bisector divides $BC$ in the ratio $AB:AC$.

Answer 1

The Correct Option is B

Step 1: Cross check with a direction vector
The unit vectors along $AB$ and $AC$ add to a vector along the bisector. $\vec{AB}=(-1,-1,-4)$ and $\vec{AC}=(-4,2,-4)$.
$\frac{\vec{AB}}{3\sqrt2}+\frac{\vec{AC}}{6}$ points to $D$, which lies on $BC$. Solving for the intersection with $BC$ leads to the same ratio $\sqrt2:2$ and the point $(2-3\sqrt2,\;3+3\sqrt2,\;6)$.

Final Answer:
Option (B). \[ \boxed{\text{(B)}} \]