Find Equation of Plane Containing Point & Line

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SUMMARY

The discussion focuses on finding the equation of a plane that contains the point (3,2,-3) and the line defined by the parametric equation (x,y,z) = (7,-4,5) + t (0,-2,2). Participants suggest using the point-normal form of the plane equation, ax + by + cz + d = 0, and emphasize the importance of finding a normal vector through the cross product of direction vectors derived from the given point and points on the line. The conversation clarifies that there is no distinction between "passing through" and "containing" points in the context of planes.

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  • Familiarity with the point-normal form of a plane equation
  • Knowledge of parametric equations of lines
  • Ability to solve systems of equations
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Homework Statement


Find the equation of the plane which contains the point (3,2,-3) and the line: (x,y,z) = (7,-4,5) + t (0,-2,2)

Homework Equations


Point-Normal equation?
a(x-x0)+b(y-y0)+c(z-z0) = 0

Not sure if this is related.

The Attempt at a Solution



First off:

The line (xyz) should be (7,-4,5) + t (0,-2,2) which becomes (7, -4-2t, 5 +7t). And then yea...

I'm behind :(. I've been sick for the past 2 weeks and I don't understand my friends' notes. So I'm not asking for any of you to solve this for me, but if someone could point me in the right direction and help me on interpreting the question, then that would be great.

Thanks.
 
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You can use the normal equation once you find a normal. You've got one direction vector which is parallel to the plane which is the direction vector to the line. Find another one by taking the difference between your point and any point on the line. Then use the cross product.
 


Or- just take two different values of t to find two different points on the line. You now have three points in the plane. You know how to find the plane containing three given points, don't you?
 


HallsofIvy said:
Or- just take two different values of t to find two different points on the line. You now have three points in the plane. You know how to find the plane containing three given points, don't you?

Do I use
ax + by + cz + d = 0
to get the equation of each point and solve for the system of equations?

Is there a difference between "passing through the points" and "contains the points" in terms of planes?
 


iamsmooth said:
Do I use
ax + by + cz + d = 0
to get the equation of each point and solve for the system of equations?

Is there a difference between "passing through the points" and "contains the points" in terms of planes?

If you don't know how to find the equation of a line using three points, why don't you just try to find a normal? You seemed to be ok with that.
 


iamsmooth said:
Do I use
ax + by + cz + d = 0
to get the equation of each point and solve for the system of equations?
That is certainly one possible way to do it. Notice that you get three equations to solve for four coefficients but that is okay: any multiple of the equation of the plane is also an equation of the plane. You can take anyone of a, b, c, or d to be 1, say.

But the way most people learn to find the plane containing three given points is to determine the vectors from one of the points to the other two and take the cross product of those- which leads back to the method Dick is suggesting.

Is there a difference between "passing through the points" and "contains the points" in terms of planes?
I don't know why but, geometrically, I tend to think of a line as "passing through" points and a plane as "containing" points! No, there is no difference at all.
 

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