the line l1 = P + k(Q-P)
= (2,3,0) + k(-2,-2,2)
point P lies in the xy plane. A plane perpendicular to l1 is
(2,3,0)*(-2,-2,2) = -10--> perpendicular plane is (x,y,z)*(-2,-2,2) = -10
or -2x - 2y + 2z = 10
find intersection line of this plane with the xy plane,: the xy plane is: z= 0 --> plug into other plane equation:
- 2x - 2y +2*0 = - 10
-x - y = -5
y = -x + 5 is the line equation perpendicular to the line PQ in the yx plane.
or, it you want the line in 3 dimensions:
first point is P(2,3,0), second point is (0,5,0)--> l2 = P + (second point -P)
l2 = (2,3,0) + k(-2,2,0)
The vector PQ can be written (-2, -2, 2).
We can parametrize this line as
x(t) = 2 - 2t
y(t) = 3 - 2t
z(t) = 2t
to the guy/gal who answered this question 1st..... how do u know that z = 0? cos the question specified that the line is parallel to plane Oxy, it didnt say that it intersects the xy-plane right?