Determine the equation of the plane that contains the points P = (1, -2, 0), Q = (3, 1, 4), and R = (0, -1, 2).
A) 2x - 8y + 5z = 18
B) 2x + 8y + 5z = -14
C) 2x + 8y - 5z = -14
D) 2x - 8y - 5z = 18
E) 2x + 8y + 5z = 18

Question

Quizplus · Accepted Answer

To find the equation of the plane, we first need two vectors that lie in the plane, which we can get by subtracting the coordinates of the given points. Let's use vectors PQ and PR:PQ = Q - P = (3 - 1, 1 - (-2), 4 - 0) = (2, 3, 4)PR = R - P = (0 - 1, -1 - (-2), 2 - 0) = (-1, 1, 2)Next, we find a normal vector to the plane by taking the cross product of PQ and PR:N = PQ × PR = |i    j    k |               |2    3    4|               |-1   1    2|= (3*2 - 4*1)i - (2*2 - 4*(-1))j + (2*1 - 3*(-1))k= (6 - 4)i - (4 + 4)j + (2 + 3)k= 2i - 8j + 5kSo, the normal vector N = (2, -8, 5). The equation of the plane can be written as:2(x - x_0) - 8(y - y_0) + 5(z - z_0) = 0,where (x_0, y_0, z_0) is any point on the plane. Substituting point P = (1, -2, 0), we get:2(x - 1) - 8(y + 2) + 5(z - 0) = 02x - 2 - 8y - 16 + 5z = 02x - 8y + 5z = 18Therefore, the correct equation is 2x - 8y + 5z = 18.

Determine the Equation of the Plane That Contains the Points