Finding a plane parallel to a line, knowing a point on the plane

[Telemachus]

Hi there, I got this problem to solve:

Given L:\begin{Bmatrix} x-y+z=1 & \mbox{ }& \\y+5z=0 & \mbox{ }& \end{matrix}, find the equation of the plane parallel to L that pass through the origin. Is it unic?

Well, I think its unic, cause there are not two different planes parallel to a line that pass through P(0,0,0).

I've started finding the director of L, \vec{d_L} let's call:
\pi:x-y+z=1 \pi':y+5z=0

(1,-1,1)\wedge{(0,1,5)}=(-6,-5,1)=\vec{d_L}

Lets call the plane \gamma, and \vec{n_{\gamma}}=(x_0,y_0,z_0) a normal vector of \gamma

<\vec{n_{\gamma}},\vec{d_L}>=0\Rightarrow{-6x_0-5y_0+z_0=0}

\gamma:-6x-5y+z=0, P(0,0,0)\in{\gamma}

So, that result is wrong, cause then \vec{n_{\gamma}}=d_L

How should I continue? I know that I should find a vector normal to L, and normal to \gamma, but I don't know how to proceed. Cause, for L I can find infinite normal vectors. Now I'm starting to think that the answer is not unique.

 

[HallsofIvy]

Hi there, I got this problem to solve:

Given L:\begin{Bmatrix} x-y+z=1 & \mbox{ }& \\y+5z=0 & \mbox{ }& \end{matrix}, find the equation of the plane parallel to L that pass through the origin. Is it unic?

Well, I think its unic, cause there are not two different planes parallel to a line that pass through P(0,0,0).

? There are, in fact, an infinite number of different planes, parallel to a given line, through the origin! Draw the unique line through the origin that is parallel to the given line. Any plane that contains that line and not the original line satisfies the conditions.

I've started finding the director of L, \vec{d_L} let's call:
\pi:x-y+z=1 \pi':y+5z=0

(1,-1,1)\wedge{(0,1,5)}=(-6,-5,1)=\vec{d_L}

Okay, another way to do that is to write the line as parametric equations. Using, say, z as parameter, we can write the second equation y= -5z. Then the first equation becomes x- (-5z)+ z= x+ 6z= 1 so that x= 1- 6z. That is the parametric equations are x= 1- 6t, y= -5t, z= t so the "direction vector" is (-6, -5, 1), just as you say.

Any plane through the origin can be written in the form Ax+ By+ Cz= 0. It will be parallel to the given line as long as its normal vector, (A, B, C), is perpendicular to the direction vector, (-6, -5, 1). That is, we must have -6A- 5B+ C= 0. That gives C= 6A+ 5B so that any plane Ax+ By+ (6A+ 5B)z= 0 satisfies the conditions.

Lets call the plane \gamma, and \vec{n_{\gamma}}=(x_0,y_0,z_0) a normal vector of \gamma

<\vec{n_{\gamma}},\vec{d_L}>=0\Rightarrow{-6x_0-5y_0+z_0=0}

\gamma:-6x-5y+z=0, P(0,0,0)\in{\gamma}

So, that result is wrong, cause then \vec{n_{\gamma}}=d_L

How should I continue? I know that I should find a vector normal to L, and normal to \gamma, but I don't know how to proceed. Cause, for L I can find infinite normal vectors. Now I'm starting to think that the answer is not unique.

 

[Telemachus]

Thank you very much.

Sorry for the misspelling :P

Bye there!