Find equation of a plane containing two lines

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Homework Help Overview

The problem involves finding the equation of a plane that contains two given lines represented by their parametric equations. The subject area pertains to vector geometry and the properties of planes in three-dimensional space.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of finding a normal vector to the plane using the cross product of direction vectors from the lines. There are questions regarding the handling of the parameter 't' in the equations and how it affects the final expression for the plane's equation.

Discussion Status

Some participants have provided guidance on how to approach the problem, including the use of cross products and identifying points on the lines. However, there is ongoing discussion about the implications of the parameter 't' in the context of the solution, and no consensus has been reached regarding the final form of the equation.

Contextual Notes

Participants note that the forum's policy is to assist without providing complete solutions, which influences the nature of the responses. There is also mention of a discrepancy between the poster's result and the book's answer regarding the presence of 't' in the final equation.

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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?
 
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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.
 
r_1=t(1,2,3)

r_2=t(3,1,8)

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

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

Hence the equation:

13x+y-5z=0
 
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.
 
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
 
Mark44 said:
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.
 
i see donaldos wrote r with the t outside before doing the determinant...what happens to that?
 
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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