Find the coordinates of the foot of the perpendicular and the perpendicular distance from the point P( 3, 2, 1) to the plane 2x – y + z + 1 = 0.
Find also the image of the point P in the plane.
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 ≡ 