# 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!

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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

$$\pi' = P + V^\bot \ .$$

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No, it isn't homework.. :) Thanks for your answer, can I always write a locus as $locus$=$Q + V$ with V= vector subspace solution of the homogeneous linear system of equations associated to my locus? –  sunrise Sep 2 '12 at 19:05
Yes, we can. (I'm pro-Obama, you know. :-) ) Ok, no kidding: just find the solutions of your system of linear equations in the way you prefer and separate all that is constant (and you'll get your $Q$) from the rest that will depend on some of the unknowns (which will be your $V$). –  a.r. Sep 2 '12 at 19:14
Essentially, yes. If you have a system of linear equations $AX=b$, then, changing the constants $b$ gives you parallel solution sets. –  a.r. Sep 3 '12 at 18:02
Essentially, no. The solution set of a system of linear equations $AX = b$ can always be written in its parametric form as $Q + V$, $Q$ being a particular solution of the system (i.e., $AQ =b$) and $V$ being the solution set of the homogeneous associated system $AX = 0$. Proof: $A(Q+V) = AQ + AV = b + 0 = b$. –  a.r. Sep 4 '12 at 13:30
Yes. (Essentially.) :-) –  a.r. Sep 5 '12 at 18:31

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.

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"Note that π is given as: π=Q+{v∈R4 : v⊥u1 and v⊥u2}" thanks a lot! I haven't thought about it! :) and so, $u_1$ and $u_2$ are vectors of direction of $\pi'$.. is it right? –  sunrise Sep 2 '12 at 20:37
Yes, that is right :) They even span $\pi'$, in the sense that each point in $\pi'$ is of the form $P + [\text{linear combination of$u_1,u_2$}]$. –  Feanor Sep 2 '12 at 21:00
yes! thank you very very much! :) –  sunrise Sep 3 '12 at 8:50

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 $2y+6z+4q-6=0.$

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Thanks :) So my $\pi'$ is given by two equations.. and, ok, Grassman confirms that, but if I assign 2 values to $a$ and $b$, what do I obtein? I can't obtein a particular plane because I need 2 equations.. isn't it? –  sunrise Sep 2 '12 at 17:58
I think, we need one more condition if we need to obtain a particular plane. –  lab bhattacharjee Sep 2 '12 at 18:05
Ok :) thanks a lot! –  sunrise Sep 2 '12 at 18:40