So I'm studying for my first exam and I'm stuck on this number.
Given both lines:
D1: (x,y,z) = (2,0,0) + k(0,3,0)
D2: (x,y,z) = (2,0,2) + k(0,0,1)
(a) Give the algebraic equation of the plane containing both lines.
I know that I have to start off by finding the intersection point of both lines, but I can't seem to solve it.
Any help is greatly appreciated!
The normal to this plane is normal to both lines, so the cross product of the direction vector of these lines will be the normal to the plane.
Finally, find the equations of the plane passing through each of the given points. Both of them should give the same equation, otherwise the planes are parallel.
In the given case, you should end up with a single plane that contains both lines.
THanks again Mate! You've been a big help lately, I really appreciate it.
You're welcome!
To find the intersection point of the two lines, you need to equate their x, y, and z coordinates.
Given:
D1: (x,y,z) = (2,0,0) + k(0,3,0)
D2: (x,y,z) = (2,0,2) + k(0,0,1)
For the x-coordinate:
2 = 2
For the y-coordinate:
0 + 3k = 0
For the z-coordinate:
0 = 2 + k
Now, solve each equation separately:
1. From the x-coordinates:
2 = 2
This equation is always true, so it does not provide any new information.
2. From the y-coordinates:
0 + 3k = 0
To solve this equation, subtract 0 from both sides:
3k = 0
And then solve for k:
k = 0
3. From the z-coordinates:
0 = 2 + k
Substitute the value of k we found earlier:
0 = 2 + 0
This equation is not true, meaning the lines do not intersect in the z-coordinate.
Since the lines do not intersect in all three coordinates, this means they are parallel and do not have an intersection point. Therefore, the two lines do not lie in a common plane.
Consequently, there is no algebraic equation that represents the plane containing both lines.