We say that $$B=\{\vec{u},\vec{v}\}$$ is an orthogonal basis if the vectors that form it are perpendicular. In other words, $$\vec{u}$$ and $$\vec{v}$$ form an angle of $$90^\circ$$.

$$\vec{u}=(3,0)$$, $$\vec{v}=(0,-2)$$ form an orthogonal basis since the scalar product between them is zero and this a sufficient condition to be perpendicular: $$$ \vec{u}\cdot\vec{v}=3\cdot0+0\cdot(-2)=0$$$

We say that $$B=\{\vec{u},\vec{v}\}$$ is an orthonormal basis if the vectors that form it are perpendicular and they have length $$1$$. Namely, $$\vec{u}$$ and $$\vec{v}$$ form an angle of $$90^\circ$$ and $$|\vec{u}|=1$$, $$|\vec{v}|=1$$.

$$\vec{u}=(1,0)$$, $$\vec{v}=(0,-1)$$ form an orthonormal basis since the vectors are perpendicular (its scalar product is zero) and both vectors have length $$1$$.

Perpendicular: $$\vec{u}\cdot\vec{v}=1\cdot0+0\cdot(-1)=0$$.

Unitary vectors (length 1): $$|\vec{u}|=\sqrt{1^2+0^2}=\sqrt{1}=1$$, $$|\vec{v}|=\sqrt{0^2+(-1)^2}=\sqrt{1}=1$$.