the normal of the given plane is
(3,0,-5) which is then the direction of your line
A vector equation of that line is
r = (x,y,z) = (2,9,-3) + t(3,0,-5) where t is your parameter
x = 2 + 3t ---> t = (x-2)/3
y = 9 + 0t ---> y = 9
z = -3 - 5t --> t = (z+3)/-5
so
(x-2)/3 = (z+3)/-5 , y = 9
A plane, π, has 3x-5z+3=0 as its Cartesian equation. Determine the Cartesian equation of a plane that is perpendicular to π and contains the point P(2,9,-3)
1 answer