Q.State True or False: If $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$, then the vectors $\vec{a}$ and $\vec{b}$ are orthogonal.

Square the given magnitude equation and use $|\vec{v}|^2=\vec{v}\cdot\vec{v}$:

$|\vec{a}|^2+2\,\vec{a}\cdot\vec{b}+|\vec{b}|^2=|\vec{a}|^2-2\,\vec{a}\cdot\vec{b}+|\vec{b}|^2.$ …

The idea

The quickest route is to square both sides, because the squared magnitude of a vector is a dot product: $|\vec{v}|^2=\vec{v}\cdot\vec{v}$. That turns the length condition into an algebraic one in $\vec{a}\cdot\vec{b}$.

Working it out

Both sides are non-negative, so squaring is reversible:

Expand each side:

Set them equal and cancel the common $|\vec{a}|^2$ and $|\vec{b}|^2$:

Conclusion …

Method: Turning a Magnitude Equation into a Dot-Product Condition by Squaring

Use this whenever an equation relates the magnitudes of vector sums/differences and you must extract an angle or a perpendicularity fact.

Steps

Step 1: Square both sides

Magnitudes are awkward, but $|\vec{v}|^2=\vec{v}\cdot\vec{v}$ turns them into dot products. Both sides here are non-negative, so squaring is reversible.

Step 2: Expand each squared magnitude …

Common Mistakes

Mistake 1: Cancelling magnitudes linearly instead of squaring

Why it's wrong: from $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$ you cannot cancel term-by-term; a magnitude does not distribute over addition. Correct approach: square both sides and use $|\vec{v}|^2=\vec{v}\cdot\vec{v}$.

Mistake 2: Mishandling the cross terms on expansion

Why it's wrong: forgetting that $|\vec{a}+\vec{b}|^2$ carries $+2\,\vec{a}\cdot\vec{b}$ while $|\vec{a}-\vec{b}|^2$ carries $-2\,\vec{a}\cdot\vec{b}$ loses the whole result. Correct approach: subtract to get $4\,\vec{a}\cdot\vec{b}=0$, hence $\vec{a}\cdot\vec{b}=0$. …