Find a plane perpendicular to a plane passing by point In $\mathbb R^4$ I have: $$\pi: \begin{cases} x+y-z+q+1=0 \\ 2x+3y+z-3q=0\end{cases}$$
I have to find $\pi' \bot$ $ \pi $ and passing by $P=(0,1,0,1)$. How can I do that? Thanks a lot!
 A: If this is homework, it would be nice you add the tag "homework".   :-)
Besides that, you could start writing your plane $\pi$ in the form
$$
\pi = Q + V
$$
where $Q \in \pi$ is any point on the plane, and $V \subset \mathbb{R}^4$ is the vector subspace which is the solution of the homogeneous linear system of equations associated to that of $\pi$; that is, you just delete all the constants (i.e., that $1$ in the first equation).
Then, compute $V^\bot$ and the perpendicular plane you're looking for will be
$$
\pi' = P + V^\bot  \ .
$$
A: Note that $\pi$ is given as:
$$ \pi = Q + \{ v \in \mathbb{R}^4 \ : \ v \perp u_1 \text{ and } v \perp u_2 \} $$
where $u_1 = (+1,+1,-1,1)$ and $u_2 = (2,3,1,-3)$, and $Q$ is a point and is not really relevant here. So you can give $\pi'$ as: 
$$\pi' = \{ P + t_1 u_1 + t_2 u_2 \ : \ t_1,t_2 \in \mathbb{R} \} $$
If that description is sufficient for you, you are done. If not, then you will need to compute (two, linearly independent) vectors that are orthogonal to $\pi'$.
To find vectors orthogonal to $\pi'$, you can for instance begin with any two vectors $w_1, w_2$ such that $u_1, u_2, w_1, w_2$ span $\mathbb{R}^4$. Now, do the Grahm-Schmidt orthogonalisation for these vectors, i.e.:
$$ u'_1 = u_1 \\
u'_2 = u_2 - \frac{\left< u'_1, u_2\right>}{\lVert u'_1 \rVert^2} u'_1 \\
w'_1 = w_1 - \frac{\left< u'_1, w_1\right>}{\lVert u'_1 \rVert^2} u'_1 
- \frac{\left< u'_2, w_1\right>}{\lVert u'_2 \rVert^2} u'_2\\
w'_2 = w_2 - \frac{\left< u'_1, w_2\right>}{\lVert u'_1 \rVert^2} u'_1 
- \frac{\left< u'_2, w_2\right>}{\lVert u'_2 \rVert^2} u'_2
\frac{\left< w'_1, w_2\right>}{\lVert w'_1 \rVert^2} w'_1
$$
That way $w'_1,w'_2$ will be orthogonal to $\pi'$ (nota bene: they will also span the $\pi$, i.e. you will have $\pi = \{ Q + t_1 w_1 + t_2 w_2 \ : \ t_1,t_2 \in \mathbb{R} \} $). You can now write:
$$
\pi' = P + \{ v \in \mathbb{R}^4 \ : \ v \perp w'_1 \text{ and } v \perp w'_2 \} 
$$
or equivalently:
$$
\pi' = \{ v \in \mathbb{R}^4 \ : \ \left<v,w_1\right> = \left<P,w_1\right> \text{ and } \left<v,w_2\right> = \left<P,w_2\right>\} 
$$
The last form can be easily turned into a set of equations.
A: The equation of any plane$(\pi_1)$ passing through $P(0,1,0,1)$ is $a(x-0)+b(y-1)+c(z-0)+d(q-1)=0$ where $a,b,c,d$ are indeterminate constants,
If  $\pi_1 \bot \pi $ , the sum of the product of the directional cosines will be $0$.
So, $(1)(a)+(1)(b)+(-1)(c)+(1)(d)=0\implies a+b-c+d=0$
and $(2)(a)+(3)(b)+(1)(c)+(-3)(d)=0\implies 2a+3b+c-3d=0 $
So, $2c=5a+6b$  and $2d=3a+4b$,
So, $\pi_1$ becomes $2a(x-0)+2b(y-1)+(5a+6b)(z-0)+(3a+4b)(q-1)=0$
Or, $a(2x+5z+3q-3)+b(2y+6z-2+4q-4)=0$
Or, $a(2x+5z+3q-3)+b(2y+6z+4q-6)=0$
If $ab≠0$, the equation of the plane $\pi_1$  will be $2x+5z+3q-3=0$ and $2y+6z+4q-6=0.$
