Question

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

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

Hint:

The condition \(A^2 + B^2 = C^2\) for the sum \(\vec{C} = \vec{A} + \vec{B}\) is the vector equivalent of the Pythagorean theorem. It holds true only when the vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a vector relationship \(\vec{A} + \vec{B} = \vec{C}\) and a scalar relationship between their magnitudes, \(A^2 + B^2 = C^2\). We need to find the angle \(\theta\) between vectors \(\vec{A}\) and \(\vec{B}\).

Step 2: Key Formula or Approach:
The magnitude of the resultant vector \(\vec{C} = \vec{A} + \vec{B}\) is given by the law of cosines for vectors:
\[ C = |\vec{C}| = \sqrt{A^2 + B^2 + 2AB \cos\theta} \]
Squaring both sides gives:
\[ C^2 = A^2 + B^2 + 2AB \cos\theta \]

Step 3: Detailed Explanation:
We have two expressions for \(C^2\):
1. From the magnitude of the vector sum: \(C^2 = A^2 + B^2 + 2AB \cos\theta\)
2. From the given information: \(C^2 = A^2 + B^2\)
Equating these two expressions for \(C^2\):
\[ A^2 + B^2 = A^2 + B^2 + 2AB \cos\theta \]
Subtract \(A^2 + B^2\) from both sides:
\[ 0 = 2AB \cos\theta \]
Assuming the vectors \(\vec{A}\) and \(\vec{B}\) are non-zero vectors (so their magnitudes \(A\) and \(B\) are non-zero), the only way for the product to be zero is if \(\cos\theta = 0\).
\[ \cos\theta = 0 \]
The angle \(\theta\) for which \(\cos\theta = 0\) is \(90^{\circ}\) or \(\frac{\pi}{2}\) radians.

Step 4: Final Answer:
The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^{\circ}\).