Find the vector equation of a plane that contains the following line L1: (x,y,z) = (1,2,3) + t(2,4,3) and does not have a z-intercept.
The Attempt at a Solution
This question was on the test and I think I might have gotten it wrong.
The correct answer is: (x,y,z) = (1,2,3) + t(2,4,3) + s(0,0,1)
It makes sense to me now because (0,0,1) is parallel to the z-axis. Hence, it does not intercept the z-axis.
What I did was... think about the question in Cartesian Equation form.
Ax + By + Cz + D = 0 (where A,B,C are normal to the plane)
I realized that the co-efficient should be 0 since there is no z-intercept. So Ax + By + D = 0.
So the normal would be (A, B, 0).
(2,4,3) CROSS (x, y, z) = (A, B, 0) [where (x,y,z) is the second direction vector]
I did the cross product formula and ended up with:
(4z-3y, 3x-2z, 2y-4x) which should equal to (A, B, 0)
2y-4x = 0
2y=4x --> y = 2x and value of z doesn't matter
Therefore, (2,4,3) CROSS (x, 2x, 0) = (A, B, 0)
Let x = 1
So: (2,4,3) and (1,2,0) are the direction vectors.
So the equation of the plane is: (x,y,z) = (1,2,3) + t(2,4,3) + s(1,2,0)
Could this be an acceptable answer?