let k= a+bt
then the line is:
x = 2 + 3t
y = -2 + 5t
z = a + bt and its direction vector is (3,5,b)
If the line is parallel to the plane then it must be perpendicular to the normal of the plane, that is
(4,3,-3)∙(3,5,b) = 0
12 + 15 - 3b = 0
b = 9
so z = a + 9t
No matter what the value of a, the line will be parallel to the plane.
trying to intersect the line with the plane we get
4(2+3t) + 3(-2+5t) - 3(a+9t) = 12
the t's drop out and we get a = -10/3
so if a = -10/3 the line is parallel to but also on the plane, while for any other value of a, the line will not be on the plane, but parallel to it
so possible values of k could be just 9t, or 5+9t, or 1+9t, etc
The vector 4i + 3j - 3k is normal to the given plane. (k is a unit vector here, not the unknown you are looking for, which I will call K). If the line represented by the parametric equations is parallel to the plane, it must be perpendicular to the normal to the plane.
The vector parallel to the line is defined by
t = (x-2)/3 = (y+2)/5 , with no component along the z axis. The line is in the x,y plane regardless of the value of K. Its vector components are
3i + 5j +0k
One can write an equation that requires the two lines to be perpendicular, but it will not involve your unknown K, and there will be no solution, because a line with only x and y components cannot be perpendicular to the plane defined by 4x + 3y – 3z -12 = 0
Are you sure you did not omit a term that involves t in the parametric definition of the line? Is it really z = K ?
Reiny treats your k as an unknown a + bt parametric term, so there are really two unknowns in that case. I treat k as an unknown constant. In Reiny's czse, solutions can be obtained, as he has done.
You should have been clearer about what k is supposed to represent.
I just copied the question as it was...I tried asking a friend and the teacher about k. When I tried to work it out, I assumed that the direction vector was 0, because it was not k+t or any number at all, but I am still unsure. Thanks.