Prove that the vectors $\vec{v} = \langle a,b \rangle$ and $\vec{w}= \langle c,d \rangle$ are orthogonal if and only if $ac+bd=0$ I've been racking my brain over this question for awhile now but haven't been able to make any meaningful progress. I was told as a hint that the dot product formula... $$\vec{v}\bullet\vec{w} = \Vert v \Vert \Vert w \Vert cos\theta$$ ...was not necessary which further confused me as this was the only place I could think of to start. I know that $\vec{v} \bullet \vec{w} = 0$ must be true but that's the extent of my knowledge. I've spent quite a bit of time researching this question as well and couldn't come up with anything pertinent. Any help or nods in the right direction would be appreciated. I apologize if the formatting of this post is wonky, this is my first time using this website. Thanks!

For vectors $\vec{v}=\langle a,b\rangle$ and $\vec{w}=\langle c,d\rangle$, we have that $\vec{v}\cdot\vec{w}=ac+bd$. We also have that $\vec{v}\cdot\vec{w}=\Vert\vec{v}\Vert\Vert\vec{w}\Vert\cos(\theta)$, where $\theta$ is the angle between $\vec{v}$ and $\vec{w}$. Note that $0\le\theta\le180^{\circ}$. Hence the following are equivalent:
$$ac+bd=0.$$
$$\vec{v}\cdot\vec{w}=0.$$
$$\Vert\vec{v}\Vert\Vert\vec{w}\Vert\cos(\theta)=0.$$
$$\text{Either }\Vert\vec{v}\Vert=0,\;\text{or}\;\Vert\vec{w}\Vert=0,\;\text{or}\;\cos(\theta)=0.$$
$$\text{Either }\vec{v}=\vec{0},\;\text{or}\;\vec{w}=\vec{0},\;\text{or}\;\theta=90^{\circ}.$$
What this shows is that there are three ways that we can have $ac+bd=0$. Either (1) $\vec{v}=\langle0,0\rangle$, (2) $\vec{w}=\langle0,0\rangle$, or (3) the angle between $\vec{v}$ and $\vec{w}$ is $\theta=90^{\circ}$.
