Question

Let L be the line passing through the point (3,1,−6) and perpendicular to the plane, P1​, defined by x−y+3z=2 Let P2​ be the plane defined by −2x+2y−6z=−7. Find the parametric equations of the line, L.

Answer 1

The parametric equations of the line, L, passing through the point (3,1,−6) and perpendicular to the plane defined by x−y+3z=2 are x = 3 + t, y = 1 - t, and z = -6 + 3t.

Step-by-step explanation:

To find the parametric equations of the line, L, passing through the point (3,1,−6) and perpendicular to the plane, P1, defined by x−y+3z=2, we need to determine the direction vector of the line.

Since the line is perpendicular to P1, the direction vector of the line will be normal to the plane.

The normal vector of P1 is (1, -1, 3). Therefore, the parametric equations of line L are:

x = 3 + t

y = 1 - t

z = -6 + 3t