Question

The planes x+y+3z = 6 and x-y+z=4 intersect in a line.

(a) Find the equation of the intersection line.
(b) Find a point on this line where z = 2 and a point on the line where = -2.
(c) Find a point half-way between the two points identified in (b).

Answer 1

The equation of the line of intersection between the planes is x=2-z, y=(5-z)/2, z=z.

Step-by-step explanation:

To find the equation of the line of intersection between the planes x+y+3z=6 and x-y+z=4, we can solve the system of equations formed by setting the two plane equations equal to each other. Using elimination, we can subtract the second equation from the first to eliminate x and y. This leaves us with 4y+2z=10, which simplifies to 2y+z=5. Rearranging this equation gives us y=(5-z)/2.

Substituting this expression for y into one of the original plane equations, we can solve for x in terms of z. Using the first plane equation, we can substitute y=(5-z)/2 and z=z to obtain x=(5-z)/2-3z/2+6. Simplifying this expression gives us x=(4-2z)/2, or x=2-z.

Therefore, the equation of the line of intersection between the planes is x=2-z, y=(5-z)/2, and z=z, or simply x=2-z, y=(5-z)/2, z=z.