Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have solved an exercise but I'm not sure to have solved it perfectly. Could you check it? It's very important for me..

In $\mathcal{R}^4$ I have a plane $\pi$ and a point P. I have to find the locus of Q points such that line PQ is perpendicular to $\pi$.

$$\pi:\begin{cases} 3x+y-z-q+1=0\\ -x-y+z+2q=0 \end{cases}$$

$P=(0,1,1,0)$

$P$ doesn't belong to $\pi$ because it doesn't satisfy first equation.

If $PQ$ line has to be $\perp$ to $\pi$, the vector of direction of $PQ$ must be perpendicular to spanners of the plane.

Spanners of plane are: $v_1=(1, -5, 0, -2)$ and $v_2=(0, 1, 1, 0)$

Then, $PQ \cdot v_1=0$ and $PQ\cdot v_2=0$

But I can write $PQ=OQ-OP$

and so I obtain $\begin{cases} (OQ-OP)\cdot v_1=0 \\(OQ-OP)\cdot v_2=0\end{cases}$

But inner product is a bilinear form and so I can also write:

$\begin{cases} OQ\cdot v_1 - OP\cdot v_1=0 \\OQ\cdot v_2-OP\cdot v_2=0\end{cases}$

I can calculate $OP\cdot v_1 $ and $OP\cdot v_2$ and, if $OQ=(x, y, z, q)$, I obtain: $\begin{cases} (x, y, z, q) \cdot (1, -5, 0, -2)=(0,1,1,0) \cdot (1,-5,0,-2) \\ (x,y,z,q) \cdot (0,1,1,0)=(0,1,1,0,) \cdot (0,1,1,0) \end{cases}$

$\begin{cases}x-5y-2q=-5 \\ y+z=2 \end{cases}$

I have solved this equations set and I have obtained that $OQ= t(1, 1/5, -1/5, 0) + s(0,-2/5, 2/5, 1) + (0, 1, 1, 0)$ (with $t, s$ belonging to $\mathbb {R}$).

And so I can say that my locus is given by $Q$ points having coordinates given by $t(1, 1/5, -1/5, 0) + s(0,-2/5, 2/5, 1) + (0, 1, 1, 0)$.

My locus is given by vectors perpendicular to $\pi$ and so I can say that my locus is the orthogonal complement of $\pi$, so it is a plane.

Is it all correct? Please, signal me all things that you think are wrong. Thank you!

EDIT:

I can develope the exercise in another way saying that if $PQ$ has to be perpendicular to $\pi$, $PQ$ has to be perpendicular to any vectors of $\pi$ and so $PQ$ belongs to orthogonal complement of $\pi$, so PQ is given by: $PQ=t(w_1)+s(w_2)$ where $w_1$ and $w_2$ are spanners of othogonal complement of $\pi$. Now I can obtain that $Q=P+t(w_1)+s(w_2)$. Is it correct? Thanks again

share|improve this question
    
@MTurgeon But I can write the plane as $\pi= P_0+V$ where V is the vector subspace solution of the homogeneous system associated to $\pi$ and $P_0$ is any point belonging to the plane.. isn't it? –  sunrise Sep 26 '12 at 15:26
1  
Yes, of course, this is why I removed my comment –  M Turgeon Sep 26 '12 at 15:28
1  
I cannot find an error. –  celtschk Sep 26 '12 at 15:29
    
@celtschk Thanks a lot! :) I have updated the question. Could you check it out again? Many thanks! –  sunrise Sep 27 '12 at 5:37
1  
Your edit is fine, too. –  celtschk Sep 30 '12 at 11:15
add comment

1 Answer 1

up vote 1 down vote accepted

It all seems good.

But, you jumped one step: how did you find your $v_1$ and $v_2$ spanner vectors of $\pi$?

share|improve this answer
    
    
I know that vectors of $\pi$ are perpendicular to (3, 1, -1, -1) and (-1,-1,1,2), so I solved the system associated to these two conditions.. :) I haven't written those steps because I'm sure that they are OK :) thank you! –  sunrise Sep 26 '12 at 15:39
    
Hi, I have updated the question.. could you check it out? Thank you very much! :) –  sunrise Sep 27 '12 at 5:39
1  
Still seems fine (didn't check every calculating in much detail). One more remark: when $t$ and $s$ can be arbitrary real numbers, you can rescale to avoid fractions, so instead of $(1, 1/5, -1/5, 0)$, you can use $(5,1,-1,0)$... –  Berci Sep 27 '12 at 9:43
    
Thanks Berci, my last doubt is: when I talk about spanners of plane, I'm talking of free (geometric) vectors that span the vector space associated to this affine space 2-dimensional. But $OP$ is not free vector, because it is applied.. So, I can do inner product between $OP$ and spanners of plane, only if apply spanners in O, turning them into applied vectors.. Isn't it? –  sunrise Sep 30 '12 at 15:36
add comment

Your Answer

 
discard

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.