Question

Given the Plane 5x – 2y + 4z – 9 = 0 

(i) Find the foot of the perpendicular drawn from the origin to the Plane. 

(ii) Write the vector and Cartesian equation of this perpendicular.

Answer 1

(i) The equation of the perpendicular line to the given plane 5x – 2y + 4z – 9 = 0 and

passing through the origin is

\(\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{4}=λ\\⇒\frac{x}{5}=\frac{y}{-2}=\frac{z}{4}=λ\)

Hence any point on this line is (5λ, – 2λ, 4λ). Let this point be on the given plane then 

⇒ 5(5λ) – 2( – 2λ) + 4(4λ) – 9 = 0 

⇒ λ = \(\frac{1}{5}\)

Then the foot of the perpendicular is

\((5\times\frac{1}{5},-2\times\frac{1}{5},4\times\frac{1}{5})\\⇒(1,\frac{-2}{5},\frac{4}{5})\)

Since the line is perpendicular to the Plane and passes through the point \((1,\frac{-2}{5},\frac{4}{5})\) 

The Cartesian equation is \(\frac{x}{5}=\frac{y}{-2}=\frac{z}{4}\)

The Vector equation is \(\bar{r}\)= \(\bar{0}\)+ λ(5i – 2j + 4k)