Question

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

• A. \(\vec{a}\) and \(\vec{b}\) are coincident.
• B. \(\vec{a}\) and \(\vec{b}\) are prependicular.
• C. Inclined to each other at 60°.
• D. \(\vec{a}\) and \(\vec{b}\) are parallel.

Hint:

To determine whether two vectors are perpendicular, check if their dot product is zero. The dot product \( \vec{a} \cdot \vec{b} = 0 \) implies that the vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular.

Answer 1

The Correct Option is B

The correct answer is: (B): \(\vec{a}\) and \(\vec{b}\) are perpendicular.

We are given that:

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

Step 1: Square both sides

We begin by squaring both sides of the equation to eliminate the magnitudes:

\( |\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \)

Step 2: Expand both sides

Now, expand both sides using the formula for the square of the magnitude of a vector, \( |\vec{u}|^2 = \vec{u} \cdot \vec{u} \):

\( (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \)

Expanding both sides gives:

\( \vec{a} \cdot \vec{a} + 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} - 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \)

Step 3: Simplify the equation

Canceling the common terms \( \vec{a} \cdot \vec{a} \) and \( \vec{b} \cdot \vec{b} \) from both sides, we are left with:

\( 2\vec{a} \cdot \vec{b} = -2\vec{a} \cdot \vec{b} \)

Step 4: Solve for \( \vec{a} \cdot \vec{b} \)

By simplifying this, we get:

\( 4\vec{a} \cdot \vec{b} = 0 \)

This implies:

\( \vec{a} \cdot \vec{b} = 0 \)

Conclusion:
The dot product of \( \vec{a} \) and \( \vec{b} \) is zero, which means that the vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular.