Find the equation that contains the lines:
2x + 3y + 4z = 6, x - 2y + z = 3
(2x - 1)/22 = (y + 2)/2 = (1 - z)/3
A plane (in point normal form) is defined by a point on the plane and a vector normal to it.
In general, the form of A(x - x0) + B(y - y0) + C(z - z0) = 0.
You can use the cross product to determine a normal vector given multiple lines on the plane.
The Attempt at a Solution
Step 1: Convert L1 to parametric form. Take the cross product of <2, 3, 4> and <1, -2, 1>
and that is the vector <11, 2, -7>. A point that is on both planes forming the line of intersection is (3, 0, 0). Thus L1: x = 3 + 11T, y = 2T, and z = -7T.
Step 2: Convert L2 to parametric form. x = 11T + 0.5, y = 2T -2 and z = 1 - 7T.
Since these lines are indeed parallel, we have to find another line or vector on the plane. Consider vector PQ, where point P (3, 0, 0) is on L1 and point Q (0.5, -2, 1) is on L2. Consequently, PQ is <-3.5, -2, 1>. The normal vector is found by taking the cross product of PQ and L1, and I am having a difficult time obtaining the correct answer, 24x - 13y + 34z = 72.