Answer :

Given :


Equation of plane :


A = (1, 1, 2)


To Find :


i) Length of perpendicular = d


ii) coordinates of the foot of the perpendicular


Formulae :


1) Unit Vector :


Let be any vector


Then unit vector of is



Where,


2) Length of perpendicular :


The length of the perpendicular from point A with position vector to the plane is given by,



Note :


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



Given equation of the plane is


………..eq(1)



As


Therefore equation of plane is


2x – 2y + 4z = -5 ……… eq(2)


From eq(1) normal vector of the plane is






From eq(1), p = -5


Given point A = (1, 1, 2)


Position vector of A is



Now,



= (1×2) + (1×(-2)) + (2×4)


= 2 – 2 + 8


= 8


Length of the perpendicular from point A to the plane is








Let P be the foot of perpendicular drawn from point A to the given plane,


Let P = (x, y, z)



As normal vector and are parallel



x = 2k+1, y = -2k+1, z = 4k+2


As point P lies on the plane, we can write


2(2k+1) – 2(-2k+1) + 4(4k+2) = -5


4k + 2 + 4k – 2 + 16k + 8 = -5


24k = -13



,




Therefore co-ordinates of the foot of perpendicular are


P(x, y, z) =


P ≡


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