Question:
In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively \(3\hat i+\hat j-\hat k\), \(-\hat i+3\hat j+p\hat k\) and \(5\hat i+q\hat j-4\hat k\), then the point \((p, q)\) lies on a line :
In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively \(3\hat i+\hat j-\hat k\), \(-\hat i+3\hat j+p\hat k\) and \(5\hat i+q\hat j-4\hat k\), then the point \((p, q)\) lies on a line :
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Right angle at A means \(\vec{AB}\cdot\vec{AC}=0\); this gives a linear relation between p and q.
Updated On: Nov 11, 2026
- parallel to \(x\)-axis.
- parallel to \(y\)-axis.
- making an acute angle with the positive direction of \(x\)-axis.
- making an obtuse angle with the positive direction of \(x\)-axis.
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The Correct Option is C
Solution and Explanation
\(\vec{AB}=(-1-3,\ 3-1,\ p+1)=(-4,2,p+1)\) and \(\vec{AC}=(5-3,\ q-1,\ -4+1)=(2,q-1,-3)\).
Right angle at A: \(\vec{AB}\cdot\vec{AC}=0\Rightarrow -8+2(q-1)-3(p+1)=0\Rightarrow 2q-3p-13=0\).
So \((p,q)\) lies on \(2y-3x=13\), i.e. \(y=\frac32x+\frac{13}{2}\), with slope \(\frac32\gt0\).
A positive slope means an acute angle with the positive x-axis.
Answer: (3) making an acute angle with the positive direction of \(x\)-axis.
Right angle at A: \(\vec{AB}\cdot\vec{AC}=0\Rightarrow -8+2(q-1)-3(p+1)=0\Rightarrow 2q-3p-13=0\).
So \((p,q)\) lies on \(2y-3x=13\), i.e. \(y=\frac32x+\frac{13}{2}\), with slope \(\frac32\gt0\).
A positive slope means an acute angle with the positive x-axis.
Answer: (3) making an acute angle with the positive direction of \(x\)-axis.
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