Find the equation of the plane that passes through P(1,0,2), Q(2,1,0), and R(3,-1,1).
To find the equation of the plane that passes through the given points P(1,0,2), Q(2,1,0), and R(3,-1,1), we first need to find two vectors that lie in the plane. We can do this by subtracting the coordinates of one point from the coordinates of the other two points to obtain two vectors that lie in the plane. Let's use vectors PQ and PR:
PQ = Q - P = (2-1, 1-0, 0-2) = (1,1,-2)
PR = R - P = (3-1, -1-0, 1-2) = (2,-1,-1)
Now, we can find the normal vector to the plane by taking the cross product of PQ and PR:
N = PQ x PR =
| i j k |
| 1 1 -2 |
| 2 -1 -1 |
= i(1*-1 - 1*-1) - j(1*-2 - 2*-1) + k(1*-1 - 2*-1)
= -2i - 3j - 1k
Now we have a normal vector to the plane, and we can use it to write the equation of the plane in point-normal form:
-2x - 3y - z + D = 0
where D is a constant to be determined. To find D, we substitute the coordinates of one of the points on the plane (say, P) into the equation:
-2(1) - 3(0) - (2) + D = 0
Solving for D, we get:
D = 4
So the equation of the plane is:
-2x - 3y - z + 4 = 0
Therefore, the equation of the plane that passes through the points P(1,0,2), Q(2,1,0), and R(3,-1,1) is -2x - 3y - z + 4 = 0.