Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 3 down vote accepted

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

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

share|cite|improve this answer
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.

share|cite|improve this answer
"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 $u_1,u_2$}]$. – Jakub Konieczny 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 $2y+6z+4q-6=0.$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.