# How do you find the set of parametric equations for the line in 3D described by the general equations x-y-z=-4 and x+y-5z=-12?

the line here is the line that constitutes the intersection of 2 planes as described

for the equation of a line in 3D, you need a point on the line, and a direction vector describing the path of the line from the point.

finding a point is easy. just set any of x,y or z to any value you like and solve the 2 equations that are now in 2 unknowns

here i will set z = 0 so the equations are

x-y = -4 x+y = -12

add them to get x = -8, so y = -4

for the direction of that line, which is the direction of the intersection of the 2 planes, we can first find another point on the line, this time i will set x = 0 so that the eqns are

-y - z = -4 y - 5z = -12

so the direction vector is

we're interested in the direction of this vector, not its scalar magnitude so we can simplify as we see fit

the line therefore is

we can also instead take the vector cross product of the normal vectors of the 2 intersecting planes

computationally that is the determinant of the following matrix

again, the line is

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