(i) 2x + 3y + 4z - 12 = 0

Given :

Equation of plane : 2x + 3y + 4z + 12 = 0

To Find :

coordinates of the foot of the perpendicular

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

From the given equation of the plane

2x + 3y + 4z – 12 = 0

⇒ 2x + 3y + 4z = 12

Direction ratios of the vector normal to the plane are (2, 3, 4)

Let, P = (x, y, z) be the foot of perpendicular perpendicular drawn from origin to the plane.

Therefore perpendicular drawn is

Let direction ratios of are (x, y, z)

As normal vector and are parallel

⇒x = 2k, y = 3k, z = 4k

As point P lies on the plane, we can write

2(2k) + 3(3k) + 4(4k) = 12

⇒ 4k + 9k + 16k = 12

⇒ 29k = 12

,

Therefore co-ordinates of the foot of perpendicular are

P(x, y, z) =

P =

(ii) Given :

Equation of plane : 5y + 8 = 0

To Find :

coordinates of the foot of the perpendicular

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

From the given equation of the plane

5y + 8 = 0

⇒ 5y = - 8

Direction ratios of the vector normal to the plane are (0, 5, 0)

Let, P = (x, y, z) be the foot of perpendicular perpendicular drawn from origin to the plane.

Therefore perpendicular drawn is

Let direction ratios of are (x, y, z)

As normal vector and are parallel

⇒x = 0, y = 5k, z = 0

As point P lies on the plane, we can write

5(5k) = -8

⇒ 25k = -8

,

Therefore co-ordinates of the foot of perpendicular are

P(x, y, z) =

P =