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

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.