What must be the mutual position of vectors a and b so that…

What must be the mutual position of vectors $\vec{a}$ and $\vec{b}$ so that: $1^\circ\quad |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \qquad\quad 2^\circ\quad |\vec{a}|+|\vec{b}|=|\vec{a}-\vec{b}|$ $3^\circ\quad |\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|$.

Step 1: Case $1^\circ$: $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$

Square both sides: \[|\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2\]

\[ 4\,\vec{a}\cdot\vec{b} = 0 \quad\Longrightarrow\quad \vec{a}\cdot\vec{b} = 0 \]

Since $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0$: $\cos\theta = 0 \quad\Longrightarrow\quad \theta = 90^\circ$.

Geometric interpretation: the diagonals $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ of the parallelogram built on $\vec{a}$ and $\vec{b}$ are equal in length, so the parallelogram is a rectangle.

\[\vec{a} \perp \vec{b}\]

Step 2: Case $2^\circ$: $|\vec{a}|+|\vec{b}|=|\vec{a}-\vec{b}|$

Square both sides: \[|\vec{a}|^2 + 2|\vec{a}||\vec{b}| + |\vec{b}|^2 = |\vec{a}|^2 - 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2\]

\[ |\vec{a}||\vec{b}| = -\vec{a}\cdot\vec{b} = -|\vec{a}||\vec{b}|\cos\theta \quad\Longrightarrow\quad \cos\theta = -1 \quad\Longrightarrow\quad \theta = 180^\circ. \]

Geometric interpretation: the vectors are antiparallel (opposite directions); subtracting an antiparallel vector adds the magnitudes, so $|\vec{a} - \vec{b}| = |\vec{a}| + |\vec{b}|$.

\[\vec{a} \uparrow\downarrow \vec{b}\]

Step 3: Case $3^\circ$: $|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|$

Square both sides: \[|\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 = |\vec{a}|^2 + 2|\vec{a}||\vec{b}| + |\vec{b}|^2\]

\[ |\vec{a}||\vec{b}|\cos\theta = |\vec{a}||\vec{b}| \quad\Longrightarrow\quad \cos\theta = 1 \quad\Longrightarrow\quad \theta = 0^\circ. \]

Geometric interpretation: equality in the triangle inequality holds if and only if the vectors point in the same direction; the parallelogram degenerates into a straight segment.

\[\vec{a} \uparrow\uparrow \vec{b}\]