Equation of a plane containing two lines

  • Thread starter Whiz
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  • #1
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Homework Statement



Find an equation for the plane containing the lines:
x-3= 4t , y-4 = t, z-1 = 0
x+1 = 12t, y-7 = 6t, z-5 = 3t

The Attempt at a Solution



I found the equations of both lines.
First line:
(3, 4, 1) + t(4, 1, 0)
(-1, 7, 5) + t(12, 6, 3)

Now I take the cross product of (4, 1, 0) and (12, 6, 3) to get (3, -12, 12)

I found two answers from reviewing this question and I'm confusing myself, so I was wondering which answer is right. (If any)

3x - 12y + 12z = 27
or
3x - 12y + 12z = 0

Thanks in advance.
 

Answers and Replies

  • #2
LCKurtz
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Well, you know several points on the plane. Try a few in your equations.
 
  • #3
Dick
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I don't think either one is right. You are right that the plane equation is 3x - 12y + 12z=C. Find the correct value for C by putting say t=0 into your line equations and solving for it.
 

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