Find the length and the foot of perpendicular from the point 
to the plane 2x – 2y + 4z + 5 = 0.
It is given that,
The point is 
.
The plane is 2x – 2y + 4z + 5 = 0
We need to find the foot of the perpendicular from the Point P to the equation of the given plane.
Also, we need to find the distance from the point P to the plane.
Let Q be the foot of the perpendicular from the point 
.
Let the point Q be Q(x1, y1, z1).
We know the direction ratio of any line segment PQ, where P(x1, y1, z1) and Q(x2, y2, z2), is given by (x2 – x1, y2 – y1, z2 – z1).
So, the direction ratio of PQ is given by
Direction ratio of PQ = (x1 – 1, y1 – 3/2, z1 – 2)
Now, let 
be the normal to the plane 2x – 2y + 4z + 5 = 0.
is obviously parallel to the 
since, a normal is an object such as a line or vector that is perpendicular to a given object.
Direction ratio simply states the number of units to move along each axis.
For any plane, ax + by + cz = d
a, b and c are normal vectors to the plane.
And therefore, the direction ratios are (a, b, c).
So, direction ratio of 
= (2, -2, 4) for plane 2x – 2y + 4z + 5 = 0.
Cartesian equation of line PQ, where P(1, 3/2, 2) and Q(x1, y1, z1) is
Let us find any point on this line.
From 
⇒ x1 – 1 = 2λ
⇒ x1 = 2λ + 1
From 
From 
⇒ z1 – 2 = 4λ
⇒ z1 = 4λ + 2
Any point on the line is (2λ + 1, 3/2 – 2λ, 4λ + 2).
This point is Q.
Q(x1, y1, z1) = Q(2λ + 1, 3/2 – 2λ, 4λ + 2) …(i)
And it was assumed to be lying on the given plane. So, substitute x1, y1 and z1 in the plane equation, we get
2x1 – 2y1 + 4z1 + 5 = 0
Simplifying it to find the value of λ.
⇒ 4λ + 2 – 3 + 4λ + 16λ + 8 + 5 = 0
⇒ 4λ + 4λ + 16λ + 2 – 3 + 8 + 5 = 0
⇒ 24λ + 12 = 0
⇒ 24λ = -12
Since, Q is the foot of the perpendicular from the point P.
The, substitute the value of λ in equation (i),
Also, we need to find 
.
Where,
P = (1, 3/2, 2)
Q = (0, 5/2, 0)
is found as,
Thus, foot of the perpendicular from the given point to the plane is (0, 5/2,0) and the distance is √6 units.