Answer :

(i) 2x + 3y + 4z - 12 = 0


Given :


Equation of plane : 2x + 3y + 4z + 12 = 0


To Find :


coordinates of the foot of the perpendicular


Note :


If two vectors with direction ratios (a1, a2, a3) & (b1, b2, b3) are parallel then



From the given equation of the plane


2x + 3y + 4z – 12 = 0


2x + 3y + 4z = 12


Direction ratios of the vector normal to the plane are (2, 3, 4)


Let, P = (x, y, z) be the foot of perpendicular perpendicular drawn from origin to the plane.


Therefore perpendicular drawn is



Let direction ratios of are (x, y, z)


As normal vector and are parallel



x = 2k, y = 3k, z = 4k


As point P lies on the plane, we can write


2(2k) + 3(3k) + 4(4k) = 12


4k + 9k + 16k = 12


29k = 12



,




Therefore co-ordinates of the foot of perpendicular are


P(x, y, z) =


P =


(ii) Given :


Equation of plane : 5y + 8 = 0


To Find :


coordinates of the foot of the perpendicular


Note :


If two vectors with direction ratios (a1, a2, a3) & (b1, b2, b3) are parallel then



From the given equation of the plane


5y + 8 = 0


5y = - 8


Direction ratios of the vector normal to the plane are (0, 5, 0)


Let, P = (x, y, z) be the foot of perpendicular perpendicular drawn from origin to the plane.


Therefore perpendicular drawn is



Let direction ratios of are (x, y, z)


As normal vector and are parallel



x = 0, y = 5k, z = 0


As point P lies on the plane, we can write


5(5k) = -8


25k = -8



,




Therefore co-ordinates of the foot of perpendicular are


P(x, y, z) =


P =


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