Question

The magnitudes of three vectors \(\vec A,\vec B,\vec C\) are \(12,5\), and \(13\) units respectively and \(\vec A+\vec B=\vec C\). The angle between \(\vec A\) and \(\vec B\) is

• A. \(0^\circ\)
• B. \(120^\circ\)
• C. \(90^\circ\)
• D. \(45^\circ\)

Hint:

If \(A^2+B^2=C^2\), then the vectors \(A\) and \(B\) are perpendicular.

Answer 1

The Correct Option is C

Concept: If: \[ \vec A+\vec B=\vec C \] then: \[ C^2=A^2+B^2+2AB\cos\theta \]

Step 1: Given: \[ A=12,\qquad B=5,\qquad C=13 \]

Step 2: Use the resultant formula. \[ 13^2=12^2+5^2+2(12)(5)\cos\theta \] \[ 169=144+25+120\cos\theta \] \[ 169=169+120\cos\theta \]

Step 3: Simplify. \[ 120\cos\theta=0 \] \[ \cos\theta=0 \]

Step 4: Therefore: \[ \theta=90^\circ \] Hence, \[ \boxed{90^\circ} \]