1.

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

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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