In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a)z = 2 (b) (c) (d)5y+ 8 = 0

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(a) The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1) The direction ratios of normal are 0, 0, and 1. ∴ Dividing both sides of equation (1) by 1, we obtain This is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance...
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(a)The equation of the plane isz= 2 or 0x+ 0y+z= 2 … (1) The direction ratios of normal are 0, 0, and 1. ∴ Dividing both sides of equation (1) by 1, we obtain This is of the formlx+my+nz=d, wherel,m,nare the direction cosines of normal to the plane anddis the distance of the perpendicular drawn from the origin. Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units. (b)x+y+z= 1 … (1) The direction ratios of normal are 1, 1, and 1. ∴ Dividing both sides of equation (1) by, we obtain This equation is of the formlx+my+nz=d, wherel,m,nare the direction cosines of normal to the plane anddis the distance of normal from the origin. Therefore, the direction cosines of the normal areand the distance of normal from the origin isunits. (c)2x+ 3y­−z= 5 … (1) The direction ratios of normal are 2, 3, and −1. Dividing both sides of equation (1) by, we obtain This equation is of the formlx+my+nz=d, wherel,m,nare the direction cosines of normal to the plane anddis the distance of normal from the origin. Therefore, the direction cosines of the normal to the plane areand the distance of normal from the origin isunits. (d)5y+ 8 = 0 ⇒ 0x− 5y+ 0z= 8 … (1) The direction ratios of normal are 0, −5, and 0. Dividing both sides of equation (1) by 5, we obtain This equation is of the formlx+my+nz=d, wherel,m,nare the direction cosines of normal to the plane anddis the distance of normal from the origin. Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin isunits. read less
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