Answer :

Given :


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


P = (1, 2, 4)


To Find :


i) Equation of perpendicular


ii) Length of perpendicular = d


iii) 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


2x + y – 2z + 3 = 0


2x + y – 2z = -3 ………..eq(1)


From eq(1) direction ratios of normal vector of the plane are


(2, 1, -2)


Therefore, equation of normal vector is






= 3


From eq(1), p = -3


Given point P = (1, 2, 4)


Position vector of A is



Here,


Now,



= (1×2) + (2×1) + (4×(-2))


= 2 + 2 - 8


= -4


Length of the perpendicular from point A to the plane is





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


Let Q = (x, y, z)



As normal vector and are parallel, we can write,



This is the equation of perpendicular.



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


As point Q lies on the plane, we can write


2(2k+1) + (k+2) - 2(-2k+4) = -3


4k + 2 + k + 2 + 4k - 8 = -3


9k = 1



,




Therefore co-ordinates of the foot of perpendicular are


Q(x, y, z) =


Q ≡


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