Find the equation for the plane through P0(-9 6, -5) perpendicular to the following line?

Question

x=-9-t

y=6+4t

z=5t

-infinity < t < infinity

using a coefficient of -1 for x the equations of the plane is __________

Answer 1

P₀(- 9, 6, - 5). Direction vector of line is < -1, 4, 5>, This vector is normal vector for unknown plane.

So, equation of plane: -1(x - (-9)) + 4(y - 6) + 5(z - (-5)) = 0; - x - 9 + 4y - 24 + 5z + 25 = 0;

-x + 4y + 5z - 8 = 0