Question

Let \( PQR \) be a triangle such that \[ \vec{PQ}=-2\hat i-\hat j+2\hat k,\quad \vec{PR}=a\hat i+b\hat j-4\hat k,\ a,b\in\mathbb{Z}. \] Let \( S \) be the point on \( QR \) which is equidistant from the lines \( PQ \) and \( PR \). If \[ |\vec{PR}|=9 \quad \text{and} \quad \vec{PS}=\hat i-7\hat j+2\hat k, \] then the value of \( 3a-4b \) is:

Hint:

For equidistance from two lines through a point, always apply the angle-bisector condition using direction vectors.

Answer 1

Correct Answer: 29

Concept: A point equidistant from two intersecting lines lies on the angle bisector. Hence, vectors \( \vec{PS} \) must satisfy proportionality with unit direction vectors of \( PQ \) and \( PR \).
Step 1: Use the magnitude condition \[ |\vec{PR}|=\sqrt{a^2+b^2+16}=9 \Rightarrow a^2+b^2=65 \]
Step 2: Use angle bisector condition \[ \frac{\vec{PS}\cdot\vec{PQ}}{|\vec{PQ}|} =\frac{\vec{PS}\cdot\vec{PR}}{|\vec{PR}|} \] Substitute vectors and simplify to obtain: \[ a=7,\quad b=4 \]
Step 3: Compute the required expression \[ 3a-4b=21-16=5 \] Final Answer: \[ \boxed{5} \]