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                                    The direction cosines of the normal to the plane 5y + 4 = 0 are A. 0,\(\frac{-4}5\),0B. 0, 1, 0 C. 0, -1, 0 D. None of these | 
                            
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Answer»  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, - y = 4/5 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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