Question

If \(|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|\) then the angle between vectors \(\vec{A}\) and \(\vec{B}\) is:

• A. 0
• B. \(\pi/4\)
• C. \(\pi/2\)
• D. \(3\pi/4\)

Hint:

Geometrically, the vectors \(\vec{A}+\vec{B}\) and \(\vec{A}-\vec{B}\) represent the diagonals of a parallelogram formed by vectors \(\vec{A}\) and \(\vec{B}\). The condition that the diagonals have equal length means the parallelogram must be a rectangle, which implies that \(\vec{A}\) and \(\vec{B}\) are perpendicular.

Answer 1

The Correct Option is C

Step 1: Square both sides of the given equation. The magnitude of a vector is always non-negative, so we can square both sides without loss of information. \[ |\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2 \]
Step 2: Express the square of the magnitude using the dot product. Recall that for any vector \(\vec{V}\), \(|\vec{V}|^2 = \vec{V} \cdot \vec{V}\). \[ (\vec{A}+\vec{B}) \cdot (\vec{A}+\vec{B}) = (\vec{A}-\vec{B}) \cdot (\vec{A}-\vec{B}) \]
Step 3: Expand the dot products. \[ \vec{A}\cdot\vec{A} + \vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{A} + \vec{B}\cdot\vec{B} = \vec{A}\cdot\vec{A} - \vec{A}\cdot\vec{B} - \vec{B}\cdot\vec{A} + \vec{B}\cdot\vec{B} \] Since \(\vec{A}\cdot\vec{A} = |\vec{A}|^2\) and the dot product is commutative (\(\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}\)): \[ |\vec{A}|^2 + 2(\vec{A}\cdot\vec{B}) + |\vec{B}|^2 = |\vec{A}|^2 - 2(\vec{A}\cdot\vec{B}) + |\vec{B}|^2 \]
Step 4: Simplify the equation to find the condition on the dot product. Canceling the \(|\vec{A}|^2\) and \(|\vec{B}|^2\) terms from both sides: \[ 2(\vec{A}\cdot\vec{B}) = -2(\vec{A}\cdot\vec{B}) \] \[ 4(\vec{A}\cdot\vec{B}) = 0 \] \[ \vec{A}\cdot\vec{B} = 0 \]
Step 5: Interpret the result. The dot product of two non-zero vectors is zero if and only if they are orthogonal (perpendicular). The angle between them is 90 degrees, or \(\pi/2\) radians.