Find a orthogonal vector of an another

Question

I need to found a orthogonal vector of an another vector that is $u=5i-3j$. I think that are multiples solutions but I’m not sure.

Answer 1

In $2 $D,

A vector which is orthogonal to $ u=a\vec{i}+b\vec{j} $ can be taken as

$ v=b\vec{i}-a\vec{j}$ .

In $3 $D,

A vector which is orthogonal to $ u=a\vec{i}+b\vec{j}+c\vec{k} $ can be as

$v= b\vec{i}-a\vec{j}$.

Answer 2

For a given Vector $\vec u=5i-3j$ , there are infinitely many vectors orthogonal to it. If we take a vector $\vec v=ai+bj$ ,we just have to make sure that $\vec u\cdot\vec k=0$ which leads to the condition: $5a=3b$.
$\vec v $ could be $\lambda(3i+5j) $ where the value of $\lambda$ is to be found with other information we can gather from the question.

Answer 3

Say we have an $n$-dimensional column vector $v$. First construct the matrix $P = \frac{vv^T}{v^Tv}$ where the numerator is an $n$ by $n$ matrix (of rank one) and the denominator is the usual dot product of $v$ with itself. $P$ has a few nice properties which I'll leave for you to verify. Firstly, $P^2=P$ so $P$ is a projection. Also, $P=P^T$ so it is symmetric. Finally, $Pw=\lambda v$ for some scalar $\lambda$, so it projects onto the span of $v$ and in paticular that $Pv=v$.

Now take some random vector $w$ that isn't in the span of $v$,which is to say $w-Pw \neq 0$. I claim this vector $w-Pw$ is orthogonal to $v$. To see this we need to show the dot product of the two vector is zero, which is to say that $(w-Pw)^Tv=0$ so we calculate $$(w-Pw)^Tv=w^T(I-P)^Tv=w^T(I-P)v=w^T(v-Pv)=w^T(v-v)=0$$

And so they are indeed orthogonal.