Question

If $|A+B| = |A-B|$ then the angle between the two vectors $A$ and $B$ is

• A. $0^{\circ}$
• B. $180^{\circ}$
• C. $120^{\circ}$
• D. $90^{\circ}$

Hint:

Geometrically, this means the diagonals of a parallelogram are equal, which only happens if the parallelogram is a rectangle (angle = $90^{\circ}$).

Answer 1

The Correct Option is D

Step 1: Concept
Use the formula for the magnitude of the sum and difference of two vectors: $|A \pm B| = \sqrt{A^2 + B^2 \pm 2AB\cos\theta}$.

Step 2: Meaning
Squaring both sides of $|A+B| = |A-B|$ simplifies the expression and removes the radical.

Step 3: Analysis
$A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 - 2AB\cos\theta$. This simplifies to $4AB\cos\theta = 0$.

Step 4: Conclusion
Since $A$ and $B$ are non-zero vectors, $\cos\theta = 0$, which means $\theta = 90^{\circ}$.
Final Answer: (D)