Cartesian equation of straight line in three dimensions?
Can u help me with the following question....
Find the Cartesian form of the equations of the straight line, l1, which passes through the points P(2,3,0) and Q(0,1,2).
Find also the Cartesian form of the equations of a straight line, l2, which passes through P, is perpendicular to l1 and parallel to the plane Oxy.
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?