Question

The plane through the point (1,0,-3) and perpendicular to the line x=1-2t, y=5-4t, z=9-2t intersects at:

a) A line
b) A point
c) A plane
d) No intersection

Answer 1

The plane through the point (1,0,-3) and perpendicular to the line x=1-2t, y=5-4t, z=9-2t intersects at a plane.

Step-by-step explanation:

The plane through the point (1,0,-3) and perpendicular to the line x=1-2t, y=5-4t, z=9-2t can be found by finding a vector that is perpendicular to the line and then using the point-normal form of the equation of a plane. The direction vector of the line is (-2,-4,-2), so a vector perpendicular to the line is (2,-1,0) or any scalar multiple of it. Using the point (1,0,-3), the equation of the plane is 2(x-1)-(y-0)+0(z-(-3))=0, which simplifies to 2x-y+6=0.

This is the equation of a plane. Therefore, the answer is c) A plane.