Question

Find an equation of the plane that passes through the point (3, 6, -1) and contains the line x = 4 - t, y = 2t - 1, z = -3t?

Answer 1

To find the equation of the plane that passes through a given point and contains a given line, calculate two direction vectors parallel to the line, take the cross product of those vectors to find the normal vector to the plane, and use the point and the normal vector to write the equation of the plane.

Step-by-step explanation:

To find an equation of the plane that passes through the point (3, 6, -1) and contains the line x = 4 - t, y = 2t - 1, z = -3t, we need to find the normal vector of the plane. The normal vector is perpendicular to the plane and can be found by taking the cross product of two vectors parallel to the plane. Using the direction vectors of the line, we can calculate the normal vector and the equation of the plane.

1. Calculate two direction vectors parallel to the plane by using the coefficients of t in the equations of the line. Let's call these vectors v1 and v2.
2. Take the cross product of v1 and v2 to find the normal vector to the plane.
3. Use the point (3, 6, -1) and the normal vector to write the equation of the plane in the form Ax + By + Cz + D = 0, where A, B, C, and D are constants.

The equation of the plane is 2x + 3y - 4z + 13 = 0.