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.

Another way to find a vector $\vec{v}$ for a given $\vec{u}$ such that
$$
\vec{u}\cdot\vec{v}=0
$$
is to use an antisymmetric matrix $A$ ($A^\top=-A$) defined as follow
$$
A_{ij}u_iu_j=0\qquad(\text{sum over }ij).
$$
In two dimension $A$ is
$$
A=\begin{pmatrix}
0&1\\
-1&0\\
\end{pmatrix}.
$$
In three dimension $A$ is
$$
A=\begin{pmatrix}
0&1&1\\
-1&0&1\\
-1&-1&0\\
\end{pmatrix}.
$$
In 2D only one such vector $\vec{v}=A\vec{u}$ exist, while in 3D you can apply the same matrix to the sum $\vec{u}+\vec{v}$ finding a vector perpendicular to the plane given by the other two vectors.
2D
The matrix $A$ can be calculated as follow
$$
A_{ij}u_iu_j=A_{11}u_1^2+(A_{12}+A_{21})u_1u_2+A_{22}u_2^2.
$$
One way is to set $A_{11}=0=A_{22}$ and $A_{21}=-A_{12}$.
3D
Again
$$
A_{ij}u_iu_j=A_{11}u_1^2+(A_{12}+A_{21})u_1u_2+A_{22}u_2^2+(A_{13}+A_{31})u_1u_3+(A_{23}+A_{32})u_2u_3+A_{33}u_3^2,
$$
and setting $A_{11}=A_{22}=A_{33}=0$ and $A_{21}=-A_{12}$, $A_{31}=-A_{13}$ and $A_{23}=-A_{32}$.
