Question

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?

Answer 1

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

Explanation:

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)]`