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

• mikee
In summary: Knowing that B = (x1, y1), C = (x2, y2), and that the line is perpendicular to the x-axis, you can use those values to find the x and y coordinates of the line's intersection with the x-axis. In summary, you can use the line equation to find the intersection point of two planes, and then use those coordinates to find the line of intersection with the x-axis.
mikee

## 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.

## 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?

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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## What is the equation of the plane?

The equation of the plane is 4x-3y-z-1+K(2x+4y+z-5)=0.

## What does K represent in the equation?

K represents a variable coefficient that can be used to manipulate the equation and find different planes that satisfy the given equation.

## How can I find the normal vector of the plane?

The coefficients of x, y, and z in the equation represent the components of the normal vector. Therefore, the normal vector of the plane is (4, -3, -1).

## What is the significance of the constant term in the equation?

The constant term in the equation represents the distance of the plane from the origin. A positive value indicates that the plane is above the origin, while a negative value indicates that the plane is below the origin.

## Can this equation be used to find the distance between a point and the plane?

Yes, the equation can be used to find the perpendicular distance between a point and the plane. The distance between a point (x0, y0, z0) and the plane can be calculated using the formula: d = |4x0-3y0-z0-1|/sqrt(4^2+(-3)^2+(-1)^2).

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