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 ≡