Find equation of a plane containing two lines

  • Thread starter hachi_roku
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  • #1
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Homework Statement


find the equation of the plane that contains the line r_1(t) = (t,2t,3t) and r_2(t)=(3t,t,8t)


Homework Equations





The Attempt at a Solution


i don't know where to start...my book does not have an example similar. can somebody just point me to the right direction?
 

Answers and Replies

  • #2
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Each of your line equations defines a vector with the same direction as the line. Find the cross product of these vectors to get a third vector, say <a, b, c>. That will be a normal to the plane. Find a point on either of your given lines, say (x0, y0, z0).

Use the normal and the given point to write the equation of the plane as a(x - x0) + b(y - y0) + c(z - z0) = 0.
 
  • #3
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[tex]r_1=t(1,2,3)[/tex]

[tex]r_2=t(3,1,8)[/tex]

Both [tex]\vec{v}_1=\left(\begin{array}{c} 1 \\ 2 \\ 3\end{array}\right)[/tex] and [tex]\vec{v}_2=\left(\begin{array}{c} 3 \\ 1 \\ 8\end{array}\right)[/tex] are in the plane.

So [tex]\vec{n}=\vec{v}_1 \times \vec{v}_2=\left(\begin{array}{c} 13 \\ 1 \\ -5\end{array}\right)[/tex] is normal to the plane.

Hence the equation:

[tex]13x+y-5z=0[/tex]
 
  • #4
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Donaldos,
It is the policy of this forum to provide help to a poster, but not to give a complete answer to someone's problem.
 
  • #5
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ok i got the answer, but when i did the cross product my answers sitll have the t in them

my ans is 13t2x+t2y-5t2z = 0

the books answer is exactly that but without the t's
 
  • #6
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Donaldos,
It is the policy of this forum to provide help to a poster, but not to give a complete answer to someone's problem.
I'm sorry. I'll keep that in mind.
 
  • #7
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i see donaldos wrote r with the t outside before doing the determinant...what happens to that?
 
  • #8
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All you need is any old vector that is parallel to the line, so any multiple of the vector will still be parallel. The t multiplier can be any real value, so it's convenient to let t = 1.
 

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