Answer :

Given :


Equation of plane : 2x - y + z + 1 = 0


P = (3, 2, 1)


To Find :


i) Length of perpendicular = d


ii) coordinates of the foot of the perpendicular


iii) Image of point P in the plane.


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 + z + 1 = 0


2x - y + z = -1 ………..eq(1)


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


(2, -1, 1)


Therefore, equation of normal vector is






From eq(1), p = -1


Given point P = (3, 2, 1)


Position vector of A is



Here,


Now,



= (3×2) + (2×(-1)) + (1×1)


= 6 - 2 + 1


= 5


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,



x = 2k+3, y = -k+2, z = k+1


As point A lies on the plane, we can write


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


4k + 6 + k – 2 + k + 1 = -1


6k = -6



,




Therefore, co-ordinates of the foot of perpendicular are


Q(x, y, z) =


Q ≡


Let, R(a, b, c) be image of point P in the given plane.


Therefore, the power of points P and R in the given plane will be equal and opposite.


2a – b + c + 1 = - (2(3) – 2 + 1 + 1)


2a – b + c + 1 = - 6


2a – b + c = - 7 ………eq(2)


Now,


As are parallel



a = 2k+3, b = -k+2, c = k+1


substituting a, b, c in eq(2)


2(2k+3) - (-k+2) + (k+1) = -7


4k + 6 + k – 2 + k + 1 = -7


6k = -12



,




Therefore, co-ordinates of the image of P are


R(a, b, c) =


R ≡


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