Find the equation of the plane that contains the points P(1, 2, 3), Q(4, 5, 6), and R(2, -1, 0).
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To find the equation of the plane that contains the points P(1, 2, 3), Q(4, 5, 6), and R(2, -1, 0), we need to find two vectors that lie on the plane. We can use the vectors PQ and PR.
Vector PQ = Q - P = (4, 5, 6) - (1, 2, 3) = (3, 3, 3)
Vector PR = R - P = (2, -1, 0) - (1, 2, 3) = (1, -3, -3)
Now, we can find the cross product of PQ and PR to get a vector that is perpendicular to the plane.
Cross product (PQ x PR) = (3, 3, 3) x (1, -3, -3) = [(3)(-3) - (3)(-3), (3)(1) - (3)(-3), (3)(-3) - (3)(1)] = (0, 12, -12)
The general equation of a plane can be written as Ax + By + Cz + D = 0, where (A, B, C) is the vector perpendicular to the plane.
Since the vector perpendicular to the plane is (0, 12, -12), we can write the equation of the plane as:
0x + 12y - 12z + D = 0
To find D, we can substitute the coordinates of one of the given points. Let's use the coordinates of point P(1, 2, 3).
0(1) + 12(2) - 12(3) + D = 0
24 - 36 + D = 0
D = 12
Therefore, the equation of the plane that contains the points P(1, 2, 3), Q(4, 5, 6), and R(2, -1, 0) is:
12y - 12z + 12 = 0
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