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: Concept
If $\vec{A}+\vec{B}=\vec{C}$, then $C^2 = A^2 + B^2 + 2AB\cos\theta$.

Step 2: Meaning
Check if the magnitudes satisfy the Pythagorean theorem ($12, 5, 13$ is a known triplet).

Step 3: Analysis
$13^2 = 12^2 + 5^2 \implies 169 = 144 + 25 = 169$. This means $2AB\cos\theta = 0$.

Step 4: Conclusion
Since $\cos\theta = 0$, the angle between $\vec{A}$ and $\vec{B}$ is $90^{\circ}$.
Final Answer: (C)