Find Equation of Plane: 4x-3y-z-1+K(2x+4y+z-5)=0

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SUMMARY

The discussion focuses on finding the equation of a plane defined by the intersection of the planes 4x - 3y - z - 1 = 0 and 2x + 4y + z - 5 = 0, which is parallel to the x-axis. The method involves determining the line of intersection of the two planes and using a direction vector <1, 0, 0> to establish the normal vector of the desired plane. The normal vector can be derived from the equation format Ax + By + Cz = D, where A must equal 0 due to the plane's parallelism to the x-axis. The solution requires solving for the parameters B and C by substituting points from the line of intersection into the plane equation.

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Homework Statement


Hello i just had a quick question, I am asked to Find the equation of the plane that passes through the line of intersection of the planes 4x - 3y - z - 1 = 0 and 2x + 4y + z - 5 = 0 and parallel to the x - axis.

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The Attempt at a Solution


Now i no if it gave you a point say A(3,4,5) i would use those as my x y and z values to solve for K in the equation 4x-3y-z-1+K(2x+4y+z-5)=0 and therefore finding the equation of the plane, but since it doenst give me a point and just a bit of information saying that it is parralel to the x axis, would i just use the elimation method to find the parametric equations for x y and z and then with that equation solve for the parameter let's say (t) and then therefore find points x,y,z and use those to solve for K?
 
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It is giving you one direction vector <1,0,0>

and you need second direction vector which you can find by finding the line equation at which two planes intersect.
(I am not sure how to find line equation of the intersection -
but I think converting both planes to vector form and making r1 = r2 would solve the problem.
There might be easier way)
 
A plane does not HAVE a "direction vector", it is determined by a normal vector. I think what rootX meant was that you can use <1, 0, 0> as one of two vectors parallel to the plane whose cross product gives the normal vector. I think a simpler way is this: you know that any plane can be written in the form Ax+ By+ Cz= 1 where <A, B, C> is a normal vector to the plane. Saying that the plane is perpendicular to the x-axis tells you that the vector rootX mentioned, <1,0,0>, in the direction of the x-axis, is perpendicular to that: <A, B, C>. <1, 0, 0,>= A= 0.

The line of intersection of two planes can be determined by solving the the equations of the planes for two of the coordinates in terms of the third. You can then use that third coordinate as the parameter. Knowing that the entire line is in the plane, you can choose any two points on the line, put their coordinates into the equation of the plane and solve for B and C.
 
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