Question

Find an equation of the following planes.

The plane passing through the origin and containing the line x=t-1, y=2t, z=3t+4.

Answer 1

To find the equation of the plane passing through the origin and containing a given line, we need to find a point on the plane and a normal vector to the plane. Using the point-normal form of the equation of a plane, we can find the equation by substituting the values.

Step-by-step explanation:

To find the equation of the plane passing through the origin and containing the line x = t-1, y = 2t, z = 3t+4, we need to find a point on the plane and a normal vector to the plane. Since the plane passes through the origin, a point on the plane is (0,0,0). The direction ratios of the line are the coefficients of t, which are (1, 2, 3). Therefore, the coordinates of the normal vector to the plane are (1, 2, 3).

Using the point-normal form of the equation of a plane, we have:

0 = 1(x-0) + 2(y-0) + 3(z-0)

Simplifying, we get:

x + 2y + 3z = 0