Question

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

• A. $0^{\circ}$
• B. $120^{\circ}$
• C. $90^{\circ}$
• D. $45^{\circ}$

Hint:

If magnitudes follow $a^2 + b^2 = c^2$, the vectors $a$ and $b$ are perpendicular.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The equation \(\vec{A} + \vec{B} = \vec{C}\) indicates that \(\vec{C}\) is the resultant of \(\vec{A}\) and \(\vec{B}\). We can use the magnitude of the resultant formula to find the angle between the two vectors.
Step 2: Key Formula or Approach:
1. Magnitude formula: \(C^2 = A^2 + B^2 + 2AB \cos \theta\).
2. Check if the magnitudes satisfy the Pythagorean theorem.
Step 3: Detailed Explanation:
Substitute the given magnitudes: \(A = 12, B = 5, C = 13\). \[ 13^2 = 12^2 + 5^2 + 2(12)(5) \cos \theta \] \[ 169 = 144 + 25 + 120 \cos \theta \] \[ 169 = 169 + 120 \cos \theta \] Subtract 169 from both sides: \[ 0 = 120 \cos \theta \] \[ \cos \theta = 0 \implies \theta = 90^\circ \]
Step 4: Final Answer:
The angle between \(\vec{A}\) and \(\vec{B}\) is 90°.