Find the angle between vectors a=m+n and b=m-n , given that…

Find the angle between vectors $\vec{a}=\vec{m}+\vec{n}$ and $\vec{b}=\vec{m}-\vec{n}$, given that $|\vec{m}|=|\vec{n}|=2$.

Step 1: Scalar product

\[ \vec{a}\cdot\vec{b} = (\vec{m}+\vec{n})\cdot(\vec{m}-\vec{n}) = \vec m\cdot\vec m - \vec m\cdot\vec n + \vec n\cdot\vec m - \vec n\cdot\vec n = |\vec{m}|^2 - |\vec{n}|^2 = 4 - 4 = 0 \]

The middle terms cancel by commutativity.

Step 2: Angle

$\vec a$ and $\vec b$ are non-zero as long as $\vec m\neq\pm\vec n$ (if $\vec m=\vec n$ then $\vec b=\vec 0$; if $\vec m=-\vec n$ then $\vec a=\vec 0$). In that case, by Problem 43,

\[\vec a\cdot\vec b=0 \quad\Longrightarrow\quad \theta = 90^\circ\] for every angle between $\vec m$ and $\vec n$.

Step 3: Geometric meaning

$\vec m+\vec n$ and $\vec m-\vec n$ are the diagonals of the parallelogram built on $\vec m$ and $\vec n$. With $|\vec m|=|\vec n|$ this parallelogram is a rhombus, so the result is the vector proof of the theorem: the diagonals of a rhombus are perpendicular. For $|\vec m|\neq|\vec n|$ we get $\vec a\cdot\vec b=|\vec m|^2-|\vec n|^2\neq0$, so the diagonals of a non-rhombus parallelogram are never perpendicular.

Step 4: Numerical example

$\vec m=(2,0)$, $\vec n=(1,\sqrt3)$ ($\varphi=60^\circ$): $\vec a=(3,\sqrt3)$, $\vec b=(1,-\sqrt3)$, and $\vec a\cdot\vec b = 3-3 = 0$ $\checkmark$