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 ≡