Answer:
Explanation:
To find the equation of the plane passing through the given points, we can use the point-normal form of the equation of a plane. The equation can be written as:
Ax + By + Cz = D
where (A, B, C) is the normal vector to the plane and (x, y, z) are the coordinates of a point on the plane.
First, we need to find the normal vector. We can do this by taking the cross product of two vectors formed by the given points. Let's consider the vectors formed by the points (0, 1, -2) and (0, 3, 4), and (0, 1, -2) and (-1, 1, 0).
Vector 1 = (0, 3, 4) - (0, 1, -2) = (0, 2, 6)
Vector 2 = (-1, 1, 0) - (0, 1, -2) = (-1, 0, 2)
Taking the cross product of Vector 1 and Vector 2, we get the normal vector:
Normal Vector = Vector1 × Vector2 = (22 - 60, 6*(-1) - 2*(-1), 20 - 22) = (4, -4, -4)
Now, we have the normal vector (A, B, C) = (4, -4, -4). We can choose any of the given points (0, 1, -2), (0, 3, 4), or (-1, 1, 0) to substitute into the equation.
Let's use the point (0, 1, -2):
4x - 4y - 4z = D
Substituting the coordinates (0, 1, -2):
4(0) - 4(1) - 4(-2) = D
0 + 4 + 8 = D
D = 12
Therefore, the equation of the plane passing through the given points is:
4x - 4y - 4z = 12
To find the z-intercept of the plane, we set x and y to 0 in the equation:
4(0) - 4(0) - 4z = 12
-4z = 12
z = -3
Therefore, the z-intercept of the plane is -3.