To find a vector equation for the plane, we need a point on the plane and a normal vector to the plane.
The line r = (1, 2, -1) + t(3, -1, 2) has a direction vector of (3, -1, 2).
Since the plane is parallel to the z-axis, it means the normal vector to the plane is perpendicular to the z-axis. Therefore, a possible normal vector is (0, 0, 1).
We can choose a point on the line: when t = 0, r = (1, 2, -1).
So, a point on the plane is (1, 2, -1), and a normal vector to the plane is (0, 0, 1).
Therefore, a vector equation for the plane is:
(r - (1, 2, -1)) · (0, 0, 1) = 0
Simplifying, we get:
(0, 0, 1) · (x - 1, y - 2, z + 1) = 0
0 + 0 + (z + 1) = 0
z + 1 = 0
z = -1
So, the equation of the plane is z = -1.
Determine a vector equation for the plane that is parallel to the z-axis containing the line r = (1, 2, -1) + t(3, -1, 2).
1 answer