Vector Algebra
Angle Bisector — Area of Triangle
nta_pyq_2026_jan
Grade 12
Question:
Let $P$ be a point in the plane of the vectors $\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}$ such that $P$ is equidistant from the lines AB and AC. If $|\overrightarrow{AP}|=\dfrac{\sqrt{5}}{2}$, then the area of the triangle ABP is:
2
$\dfrac{\sqrt{30}}{4}$
$\dfrac{3}{2}$
$\dfrac{\sqrt{26}}{4}$
Step-by-Step Solution
Key Concept: P equidistant from AB and AC through A means P lies on the angle bisector. Unit vectors: $\hat{u}_{AB}=\frac{3\hat{i}+\hat{j}-\hat{k}}{\sqrt{11}}$, $\hat{u}_{AC}=\frac{\hat{i}-\hat{j}+3\hat{k}}{\sqrt{11}}$. Bisector direction $\propto 2\hat{i}+\hat{k}$.
Area of $\triangle ABP=\dfrac{\sqrt{30}}{4}$.
Correct Answer: 2