Question

If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), then

• A. \(\vec{a}\) is parallel to \(\vec{b}\)
• B. \(\vec{a}\) is perpendicular to \(\vec{b}\)
• C. \(|\vec{a}|=|\vec{b}|\)
• D. \(\vec{a}\) and \(\vec{b}\) are unit vectors

Hint:

If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), then \(\vec{a}\cdot\vec{b}=0\), so the vectors are perpendicular.

Answer 1

The Correct Option is B

Concept:
For vectors, we use: \[ |\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) \] and: \[ |\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b}) \] Also: \[ \vec{a}\cdot\vec{b}=0 \] means that \(\vec{a}\) and \(\vec{b}\) are perpendicular vectors.

Step 1: Start with the given condition.
Given: \[ |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \] Squaring both sides: \[ |\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2 \]

Step 2: Expand the left-hand side.
\[ |\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) \] \[ =|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2 \]

Step 3: Expand the right-hand side.
\[ |\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b}) \] \[ =|\vec{a}|^2-2\vec{a}\cdot\vec{b}+|\vec{b}|^2 \]

Step 4: Equate both expressions.
Since: \[ |\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2 \] we get: \[ |\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2 = |\vec{a}|^2-2\vec{a}\cdot\vec{b}+|\vec{b}|^2 \] Cancel common terms: \[ 2\vec{a}\cdot\vec{b}=-2\vec{a}\cdot\vec{b} \] \[ 4\vec{a}\cdot\vec{b}=0 \] \[ \vec{a}\cdot\vec{b}=0 \]

Step 5: Use the dot product condition.
Since: \[ \vec{a}\cdot\vec{b}=0 \] therefore: \[ \vec{a}\perp\vec{b} \] So, \(\vec{a}\) is perpendicular to \(\vec{b}\). Hence, the correct answer is: \[ \boxed{(B)\ \vec{a}\text{ is perpendicular to }\vec{b}} \]