Question

If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\) then the angle between \(\vec{a}\) and \(\vec{b}\) is

• A. \(0^{\circ }\)
• B. \(45^{\circ }\)
• C. \(90^{\circ }\)
• D. \(180^{\circ }\)

Hint:

Square both sides of the given modulus condition and compare.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Square both sides of \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\) to use dot products.

Step 2: Expand
\(|\vec{a}|^{2}+2\vec{a}\cdot\vec{b}+|\vec{b}|^{2}=|\vec{a}|^{2}-2\vec{a}\cdot\vec{b}+|\vec{b}|^{2}\).

Step 3: Simplify
\(4\,\vec{a}\cdot\vec{b}=0\), so \(\vec{a}\cdot\vec{b}=0\).

Step 4: Conclusion
\(|\vec{a}||\vec{b}|\cos\theta=0\) gives \(\cos\theta=0\), so \(\theta=90^{\circ}\).

Final Answer:
The angle between the vectors is 90 degrees. \[ \boxed{\text{(C) }90^{\circ}} \]