Q.Prove that two vectors $\vec a$ and $\vec b$ are perpendicular iff $|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2$.

Perpendicular Vectors Condition

Two arrows that meet at a right angle — one east, one north — are perpendicular (or orthogonal) vectors. How do you check this without a protractor, especially in 3D where the angle is hard to draw?

The Idea: Zero Overlap

When two vectors are perpendicular, neither "borrows" any length from the other: walking along one makes zero progress in the direction of the other. The tool that measures this overlap is the dot product.

Why? Using $\vec{a}\cdot\vec{b}=\lVert\vec{a}\rVert\,\lVert\vec{b}\rVert\cos\theta$, a right angle gives $\cos 90^\circ=0$, so the dot product vanishes. In coordinates, for $\vec{a}=(a_1,a_2,a_3)$ and $\vec{b}=(b_1,b_2,b_3)$,

and you simply check whether this sum is $0$.

Examples

2D: $\vec{a}=(3,4)$, $\vec{b}=(4,-3)$: $\;3(4)+4(-3)=12-12=0$ — perpendicular. (In general $(x,y)$ and $(y,-x)$ are always perpendicular.)

3D: $\vec{p}=(1,2,3)$, $\vec{q}=(2,-1,0)$: $\;2-2+0=0$ — perpendicular.

Not every pair qualifies: $(2,1)\cdot(1,3)=2+3=5\neq 0$, so those two are not perpendicular.

Why It Matters …