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)

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