*Set y = 0 and z = 0. Then the point with x = 1, namely (1,0,0) satisfies both equations, and is a point of intersection for both planes. Since planes must be parallel or else intersect in a line, this point must be on the line of intersection.
*[This step is not essential to the solution, but does establish that the planes are not parallel, and thus the line of intersection for which we seek parametric and symmetric equations does EXIST. It also gives us a point on the line of intersection which we will call (x-zero, y-zero, z-zero) = (1, 0, 0) .]
Let the direction vector (call it v) equal the cross product of the two normal vectors (call them n1 and n2) to the given planes. We do this since that cross product must be perpendicular to each of the normal vectors, hence must lie in both planes, placing it on the line of intersection of the planes. A vector ON the line of intersection can be a direction vector for that same line.
v = n1 x n2 =
| i j k |
|1 2 3 |
|1 -1 1|
= [ (2)(1)-(3)(-1) ] i-hat — [(1)(1)-(3)(1)] j-hat + [(1)(-1) -(2)(1)] k-hat
= 5 i-hat + 2 j-hat —3 k-hat = < 5, 2, -3> = <a,b,c>.
With (1,0,0) = (x-zero, y-zero, z-zero) from our first step above,
we have parametric equations:
x = x-zero + at y = y-zero + bt z = z-zero + ct
x = 1 + 5t y = [0 + ] 2t z = [ 0 + ] -3t ,
and symmetric equations:
(x-1)/5 = y/2 = [ z/ (-3)] = -z/3.
(Note that this is consistent with Mark M.’s answer, but with my z = -3t, a different, but equivalent parametrization.
There is a minor copying error in his answer, where the correct value of y obtained earlier in his response ( y = (-2/3) z) loses its negative a few lines down from that, where he has y = (2/3) t [ = (2/3) z]. So in his parametric and symmetric equations, the y-value is off by a negative sign).
David F.
09/25/21