Question

If \(A+B=C\) and \(A^2+B^2=C^2\), then the angle between vectors \(A\) and \(B\) is

• A. \(0^\circ\)
• B. \(60^\circ\)
• C. \(90^\circ\)
• D. \(120^\circ\)

Hint:

Use \(C^2=A^2+B^2+2AB\cos\theta\). If \(C^2=A^2+B^2\), then \(\cos\theta=0\), so \(\theta=90^\circ\).

Answer 1

The Correct Option is C

We are given: \[ \vec A+\vec B=\vec C. \] Now take magnitude squared on both sides: \[ |\vec C|^2=|\vec A+\vec B|^2. \] Using vector formula: \[ |\vec A+\vec B|^2=A^2+B^2+2AB\cos\theta. \] So, \[ C^2=A^2+B^2+2AB\cos\theta. \] But the question gives: \[ A^2+B^2=C^2. \] Therefore, \[ C^2=A^2+B^2. \] Compare both equations: \[ A^2+B^2=A^2+B^2+2AB\cos\theta. \] Cancel \(A^2+B^2\) from both sides: \[ 0=2AB\cos\theta. \] Since \(A\) and \(B\) are non-zero vectors: \[ \cos\theta=0. \] Therefore, \[ \theta=90^\circ. \] Hence, the angle between vectors \(A\) and \(B\) is: \[ 90^\circ. \]