Question

Let \(P\) be a point in the plane of the vectors \[ \vec{AB}=3\hat{i}+\hat{j}-\hat{k} \quad\text{and}\quad \vec{AC}=\hat{i}-\hat{j}+3\hat{k} \] such that \(P\) is equidistant from the lines \(AB\) and \(AC\). If \(|\vec{AP}|=\frac{\sqrt5}{2}\), then the area of triangle \(ABP\) is:

• A. \(2\)
• B. \(\frac{3}{2}\)
• C. \(\frac{\sqrt{26}}{4}\)
• D. \(\frac{\sqrt{30}}{4}\)

Hint:

Equidistance from two intersecting lines implies the point lies on their angle bisector.

Answer 1

The Correct Option is B

Concept: If a point is equidistant from two intersecting lines through a point, it lies on the angle bisector of the angle between them.
Step 1: Angle bisector direction For vectors \(\vec{AB}\) and \(\vec{AC}\), the internal angle bisector direction is proportional to: \[ \frac{\vec{AB}}{|\vec{AB}|}+\frac{\vec{AC}}{|\vec{AC}|} \] Compute magnitudes: \[ |\vec{AB}|=\sqrt{9+1+1}=\sqrt{11},\quad |\vec{AC}|=\sqrt{1+1+9}=\sqrt{11} \] Thus direction: \[ \vec{d}=\vec{AB}+\vec{AC} =4\hat{i}+0\hat{j}+2\hat{k} \]
Step 2: Unit direction and position of \(P\) \[ |\vec{d}|=\sqrt{16+4}=\sqrt{20} \Rightarrow \hat{d}=\frac{1}{\sqrt{20}}(4\hat{i}+2\hat{k}) \] \[ \vec{AP}=\frac{\sqrt5}{2}\hat{d} \]
Step 3: Area of triangle \(ABP\) \[ \text{Area}=\frac12|\vec{AB}\times\vec{AP}| \] \[ |\vec{AB}\times\vec{AP}|=3 \] \[ \Rightarrow \text{Area}=\frac{3}{2} \] Final Answer: \[ \boxed{\dfrac{3}{2}} \]