Mark against the correct answer in each of the following:
The direction cosines of the normal to the plane 5y + 4 = 0 are
Given: Equation of plane is 5y + 4 = 0
Formula Used: Equation of a plane is lx + my + nz = p where (l, m, n) are the direction cosines of the normal to the plane and (x, y, z) is a point on the plane and p is the distance of plane from origin.
Explanation:
Given equation is 5y = -4
Dividing by -5,
which is of the form lx + my + nz = p where l = 0, m = -1, n = 0
Therefore, direction cosines of the normal to the plane is (0, -1, 0)