Question

Let \(\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}\) and \(\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}\). If \(P\) is the point on the bisector of angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) such that \(|\overrightarrow{AP}|=\dfrac{\sqrt{5}}{2}\), then the area of \(\triangle APB\) is:

• A. \(\sqrt{30}\)
• B. \(\sqrt{15}\)
• C. \(\dfrac{\sqrt{30}}{4}\)
• D. \(\dfrac{\sqrt{15}}{4}\)

Hint:

If two vectors have equal magnitudes, the angle bisector direction is simply along their {sum}.

Answer 1

The Correct Option is C

Step 1: Magnitudes of the given vectors
\[ |\overrightarrow{AB}|=\sqrt{3^2+1^2+(-1)^2}=\sqrt{11} \] \[ |\overrightarrow{AC}|=\sqrt{1^2+(-1)^2+3^2}=\sqrt{11} \]
Step 2: Direction of angle bisector
For two vectors of equal magnitude, the angle bisector is along their sum: \[ \overrightarrow{AB}+\overrightarrow{AC} =(3+1)\hat{i}+(1-1)\hat{j}+(-1+3)\hat{k} =4\hat{i}+2\hat{k} \] Magnitude: \[ |\overrightarrow{AB}+\overrightarrow{AC}|=\sqrt{4^2+0^2+2^2}=\sqrt{20}=2\sqrt{5} \] Unit vector along bisector: \[ \hat{u}=\frac{4\hat{i}+2\hat{k}}{2\sqrt{5}} =\frac{2\hat{i}+\hat{k}}{\sqrt{5}} \]
Step 3: Position vector of point \(P\)
Given: \[ |\overrightarrow{AP}|=\frac{\sqrt{5}}{2} \] \[ \Rightarrow \overrightarrow{AP} =\frac{\sqrt{5}}{2}\cdot \hat{u} =\frac{\sqrt{5}}{2}\cdot \frac{2\hat{i}+\hat{k}}{\sqrt{5}} =\hat{i}+\frac{1}{2}\hat{k} \]
Step 4: Area of \(\triangle APB\)
\[ \text{Area}=\frac{1}{2}\left|\overrightarrow{AP}\times\overrightarrow{AB}\right| \] \[ \overrightarrow{AP}=\hat{i}+\frac{1}{2}\hat{k},\quad \overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k} \] \[ \overrightarrow{AP}\times\overrightarrow{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 0 & \tfrac12\\ 3 & 1 & -1 \end{vmatrix} = -\frac12\hat{i} +\frac52\hat{j} +\hat{k} \] Magnitude: \[ \left|\overrightarrow{AP}\times\overrightarrow{AB}\right| =\sqrt{\left(\frac12\right)^2+\left(\frac52\right)^2+1^2} =\sqrt{\frac{30}{4}} =\frac{\sqrt{30}}{2} \]
Step 5: Final area
\[ \text{Area} =\frac{1}{2}\cdot\frac{\sqrt{30}}{2} =\frac{\sqrt{30}}{4} \] \[ \boxed{\dfrac{\sqrt{30}}{4}} \]