How to find perpendicular vector to another vector? How do I find a vector perpendicular to a vector like this: $$3\mathbf{i}+4\mathbf{j}-2\mathbf{k}?$$
Could anyone explain this to me, please?
I have a solution to this when I have $3\mathbf{i}+4\mathbf{j}$, but could not solve if I have $3$ components...
When I googled, I saw the direct solution but did not find a process or method to follow. Kindly let me know the way to do it. Thanks.

One way to do this is to express the vector in terms of a spherical coordinate system. For example
$$ \boldsymbol{e}= \pmatrix{a \\ b \\ c} = r \pmatrix{ \cos\varphi \cos\psi \\ \sin\varphi \cos\psi \\ \sin\psi} $$
where $r=\sqrt{a^2+b^2+c^2}$, $\tan(\varphi) = \frac{b}{a}$ and $\tan{\psi} = \frac{c}{\sqrt{a^2+b^2}}.$
Provided that $a \neq 0$ or $b \neq 0$ then
A choice of two orthogonal vectors can be found with $$ \begin{aligned} \boldsymbol{n}_1 & = \frac{{\rm d} \boldsymbol{e}}{{\rm d} \varphi}
 = r\pmatrix{-\sin \varphi \cos\psi \\ \cos\varphi \cos\psi \\ 0}& 
\boldsymbol{n}_2 & = \frac{{\rm d} \boldsymbol{e}}{{\rm d} \psi}
 = r\pmatrix{-\cos\varphi \sin\psi \\ -\sin\varphi \sin\psi \\ \cos\psi}
\end{aligned}$$
Of course, any non-zero linear combination of these two vectors is also orthogonal
$$ \boldsymbol{n} = \cos(t) \boldsymbol{n}_1 + \sin(t) \boldsymbol{n}_2 $$
where $t$ is an rotation angle about the vector $\boldsymbol{e}$.
Put it all together to make a family of orthogonal vectors in terms of $t$ as
$$ \boldsymbol{n} = \pmatrix{-b \cos(t) - \frac{a c}{\sqrt{a^2+b^2}} \sin(t) \\ a \cos(t) - \frac{b c}{\sqrt{a^2+b^2}} \sin(t) \\ \sqrt{a^2+b^2} \sin(t)} $$
For $\boldsymbol{e} = \pmatrix{3 & 4 & -2}$ the aboves gives
$$ \boldsymbol{n} = \pmatrix{
  \frac{6}{4} \sin(t)-4 \cos(t) \\ 3 \cos(t) - \frac{8}{5} \sin(t) \\ 5 \sin(t) } \longrightarrow \begin{cases} \boldsymbol{n} = \pmatrix{-4 & 3 & 0} & t =0 \\ \boldsymbol{n} =\pmatrix{\frac{6}{5} & \frac{8}{5} & 5} & t = \frac{\pi}{2} \end{cases} $$

For the case when $a=0$ and $b=0$ then you know can assign the perpendicular somewhat arbitrarily with
$$ \boldsymbol{n}_1 = \pmatrix{1 \\ 0 \\ 0} $$
and
$$ \boldsymbol{n}_2 = \pmatrix{0 \\ 1 \\ 0} $$
for the general solution
$$\boldsymbol{n} = \cos(t) \boldsymbol{n}_1 + \sin(t) \boldsymbol{n}_2$$
