Hi there, I got this problem to solve:(adsbygoogle = window.adsbygoogle || []).push({});

Given L:[tex]\begin{Bmatrix} x-y+z=1 & \mbox{ }& \\y+5z=0 & \mbox{ }& \end{matrix}[/tex], 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, [tex]\vec{d_L}[/tex] lets call:

[tex]\pi:x-y+z=1[/tex] [tex]\pi':y+5z=0[/tex]

[tex](1,-1,1)\wedge{(0,1,5)}=(-6,-5,1)=\vec{d_L}[/tex]

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

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

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

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

How should I continue? I know that I should find a vector normal to L, and normal to [tex]\gamma[/tex], 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.

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# Homework Help: Finding a plane parallel to a line, knowing a point on the plane

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