Question

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 

(i) 2x + 3y + 4z -12 = 0 

(ii) 5y + 8 = 0

Answer 1

(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

\(\frac{a_1}{b_1}\) = \(\frac{a_2}{b_2}\) = \(\frac{a_3}{b_3}\)

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 \(\overline{OP}\).

⇒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

(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

\(\frac{a_1}{b_1}\) = \(\frac{a_2}{b_2}\) = \(\frac{a_3}{b_3}\)

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 \(\overline{OP}\)

⇒x = 0, y = 5k, z = 0 

As point P lies on the plane, we can write 

5(5k) = - 8 

⇒ 25k = - 8

z = 0

Therefore co-ordinates of the foot of perpendicular are