Question

If \(|\vec A+\vec B|=|\vec A-\vec B|\), then the angle between the two vectors \(\vec A\) and \(\vec B\) is

• A. \(0^\circ\)
• B. \(180^\circ\)
• C. \(120^\circ\)
• D. \(90^\circ\)

Hint:

If \(|\vec A+\vec B|=|\vec A-\vec B|\), then the vectors are perpendicular.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem relates the magnitude of the sum and the difference of two vectors.
Geometrically, if the diagonals of a parallelogram formed by two vectors are equal, the parallelogram is a rectangle.
Algebraically, we can square the magnitudes and use dot product properties.
Step 2: Key Formula or Approach:
The square of the magnitude of the sum: \(|\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB \cos \theta\).
The square of the magnitude of the difference: \(|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2AB \cos \theta\).
Equate these two expressions.
Step 3: Detailed Explanation:

Given: \(|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|\)

Square both sides to eliminate the implicit square roots:
\[ |\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2 \]

Using the formula for vector addition and subtraction magnitudes:
\[ A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta \]

Subtract \(A^2\) and \(B^2\) from both sides:
\[ 2AB \cos \theta = -2AB \cos \theta \]

Bring all terms to one side:
\[ 2AB \cos \theta + 2AB \cos \theta = 0 \]
\[ 4AB \cos \theta = 0 \]

Assuming the vectors \(\vec{A}\) and \(\vec{B}\) are non-zero:
\[ \cos \theta = 0 \]

The angle whose cosine is zero is \(90^\circ\).
\[ \theta = 90^\circ \]

Step 4: Final Answer:
The equality of the sum and difference magnitudes implies that the vectors are perpendicular to each other. Hence, the angle is \(90^\circ\).