Determine the Cartesian equation of the plane containing the points (0, 1, -4) and (2, -1, -7), and also parallel to the line

Question

Determine the Cartesian equation of the plane containing the points (0, 1, -4) and (2, -1, -7), and also parallel to the line L: x= -4+3s, y= -2-s, z= -5+2s

Answer 1

(2, -1, -7) - (0, 1, -4) = (2, -2, -3)

The direction of the other line is (3, -1, 2) - take the s coefficients.

Then, calculate the cross-product.

i j k

2 -2 -3

3 -1 2

-7i - 13j + 4k

Thus, our plane is of the form -7x - 13y + 4z

As it contains the point 0, 1, -4, substituting, we get -7(0) -13(1) +4(-4) = -29

Thus, our plane is -7x -13y + 4z = -29, or 7x + 13y - 4z = 29