Prove that the vectors $\vec{v} = \langle a,b \rangle$ and $\vec{w}= \langle c,d \rangle$ are orthogonal if and only if $ac+bd=0$

Question

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!

Answer 1

I think the misunderstandings will mostly stem from how we define things. At some point, you have to define what it means for two vectors $\vec{v},\vec{w}$ to be orthogonal. Since you as asked for a proof, this presumably means you defined orthogonal another way instead of what is alluded to in the comments (which is also a valid way of defining things, which is precisely what this proof is asking you to show!). So let's use your way: we say $\vec{v},\vec{w}$ are orthogonal if their Euclidean dot product, defined as $|\vec{v}||\vec{w}|\cos(\theta)$, is equal to zero. What is $\theta$? It is the angle between $\vec{v},\vec{w}$.

We define the algebraic dot product as $\vec{v}\cdot\vec{w} = v_1w_1 + v_2w_2$. We want to show these define the same concepts.

By the law of cosines, considering the vectors $\vec{v},\vec{w},\vec{v-w}$, we have a triangle formed by them, and thus have

$$|\vec{v}-\vec{w}|^2 = |\vec{v}|+|\vec{w}|^2 - 2|\vec{v}||\vec{w}|cos(\theta)$$

and since $|\vec{v}| = v_1^2 + v_2^2$, we can simplify the above to get

$$(v_1-w_1)^2 + (v_2-w_2)^2 = (v_1^2+v_2^2) + (w_1^2+v_2^2) - 2|\vec{v}||\vec{w}|cos(\theta)$$

and simplifying and re-arranging gives us

$$v_1w_1 + v_2w_2 = |\vec{v}||\vec{w}|cos(\theta)$$

and so one is equal to zero if and only if the other is, and so we are done.

Answer 2

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 $\theta=90^{\circ}$.

Answer 3

Here's another approach. I'll assume that neither $\vec{v}$ nor $\vec{w}$ is the zero vector.

Suppose first that $\vec{v} = \langle a , 0 \rangle$ and $\vec{w} = \langle c, d \rangle$. Since $\vec{v}$ is horizontal, $\vec{v}$ and $\vec{w}$ are perpendicular exactly when $\vec{w}$ is vertical, i.e. $c = 0$. Since $a$ is non-zero, we can also conclude $\vec{v}$ and $\vec{w}$ are orthogonal if and only if $ac = 0$.

Now suppose that $\vec{v} = \langle a, b \rangle$ is arbitrary. To reduce to the above case, we can rotate both vectors until $\vec{v}$ is horizontal. This sends $\vec{v}$ to the horizontal vector $\vec{v'} = \langle \| \vec{v} \|, 0 \rangle$, but where does $\vec{w}$ go? I claim that $$\vec{w'} = \langle c\frac{a}{ \| \vec{v} \|} + d\frac{b}{ \| \vec{v} \|}, c\frac{-b}{ \| \vec{v} \|} + d\frac{a}{ \| \vec{v} \|} \rangle. $$ To show this, you could write down a 2-D rotation matrix and apply it to $\vec{w}$.

Applying the logic from before, $\vec{v'}$ and $\vec{w'}$ are orthogonal precisely when $\vec{w'}$ is vertical, i.e. $$c\frac{a}{ \| \vec{v} \|} + d\frac{b}{ \| \vec{v} \|} = 0. $$

Answer 4

(1) For any two-dimensional vector $v=(v_1,v_2)$ we define $Jv=(-v_2,v_1)$. One can easily -- using congruence, e.g. -- prove that $v$ and $Jv$ are orthogonal.

(2) Observe that $\det(v,w)=\langle Jv,w\rangle$.

(3) Given the vectors $v=(a,b)$ and $w=(c,d)$ satisfying $ac+bd=0$ it follows from $$0=\langle\begin{pmatrix}a \\ b\end{pmatrix},\begin{pmatrix} c \\ d \end{pmatrix}\rangle =\langle J\begin{pmatrix}b \\ -a\end{pmatrix},\begin{pmatrix} c \\ d \end{pmatrix}\rangle= \det\begin{pmatrix}b & c\\-a & d\end{pmatrix}= \det(-Jv,w)$$ that $-Jv$ and $w$ are parallel, hence $v$ and $w$ are orthogonal.