Find the equation, in the form of ax + by + cz = d, of the plane going through a point
P = (-3, 4, 2) to a line L: (x, y, z) = (3, -2, -1) + (1, -2, 2)t.
Can you solve this with brief explanation ?
Hi eunse16.
The plane passes through so
And every point on the line lies on the plane. Take any two points on the line, say and and plug the values into the plane equation:
Solve the simultaneous equations for in terms of
You should get hence the equation of the plane is
Another way to do this: Choose two points on the line. For example, if t= 0 then (x, y, z) = (3, -2, -1) + (1, -2, 2)0= (3, -2, -1) and if t= 1, (x, y, z) = (3, -2, -1) + (1, -2, 2)1= (4, -4, 1). That gives (-3, 4, 2), (3, -2, -1), and (4, -4, 1) as three points in the plane and we can use the standard method of finding the equation of the plane that includes three given points.
Now, the vector from (-3, 4, 2) to (3, -2, -1) is 6i- 6j- 3k and the vector from (-3, 4, 2) to (4, -4, 1) is 7i- 8j- k. The cross product of those two vectors is a normal vector to the plane and its equation can be written A(x-x0)+ B(y-y0)+ C(z-z0)= 0 where Ai+ Bj+ Ck is a normal vector and (x0, y0, z0) is a point in the plane.