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?

Question

Isaiah Allen · Accepted Answer

#vec r = [(-8),(-4),(0)] + lambda [ (6),(4),(2)]#

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.

#vec r = vec r_o + lambda vec d #

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

this we have the fixed point #r_o = [(x_0),(y_o),(z_o)] = [(-8),(-4),(0)]#

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

add them to get z = 8/3, y = 4/3 so we have fixed point this we have the fixed point #vec r_1 = [(x_1),(y_1),(z_1)] = [(0),(4/3),(8/3)]#

so the direction vector is

#vec d = vec r_1 - vec r_0#

# = [(x_1),(y_1),(z_1)] =[(0),(4/3),(8/3)]- [(-8),(-4),(0)] = [(8),(16/3),(8/3)] = 1/12[(6),(4),(2)]#

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

#vec r = [(-8),(-4),(0)] + lambda [ (6),(4),(2)]#

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

for generalised plane ax + by + cz = d, the normal vector is #[(a),(b),(c)]#

so here we have #vec d = [(1),(-1),(-1)] times [(1),(1),(-5)]#

computationally that is the determinant of the following matrix

#[(hat i, hat j,hat k), (1,-1,-1), (1,1,-5)]#

= #hat i# (-1-5- (-1)1) -#hat j# (1(-5) - 1-1) + #hat k# (11 - 1(-1))

= #hat i# (6) -#hat j# (-4) + #hat k# (2)

#= [ (6),(4),(2)]#

again, the line is

#vec r = [(-8),(-4),(0)] + lambda [ (6),(4),(2)]#