If $|\overrightarrow{\mathbf{a}}|=|\overrightarrow{\mathbf{b}}|$, then necessarily it implies $\overrightarrow{\mathbf{a}}= \pm \overrightarrow{\mathbf{b}}$.

Position vector of a point $\overrightarrow{\mathbf{P}}$ is a vector whose initial point is origin.

If $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|$, then the vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are orthogonal

The formula $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})^2=\overrightarrow{\mathbf{a}}^2+\overrightarrow{\mathbf{b}}^2+2 \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$ is valid for non-zero vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$.

If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are adjacent sides of $\mathbf{a}$ rhombus, then $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0$