# Find the length and the foot of perpendicular

Question:

Find the length and the foot of perpendicular from the point (1, 3/2, 2) to the plane 2– 2+ 4+ 5 = 0.

Solution:

Given plane is 2x – 2y + 4z + 5 = 0 and point (1, 3/2, 2)

The direction ratios of the normal to the plane are 2, -2, 4

So, the equation of the line passing through (1, 3/2, 2) and direction ratios are equal to the direction ratios of the normal to the plane i.e. 2, -2, 4 is

$\frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda$

Now, any point in the plane is 2λ + 1, -2λ + 3/2, 4λ + 2

Since, the point lies in the plane, then

2(2λ + 1) – 2(-2λ + 3/2) + 4(4λ + 2) + 5 = 0

4λ + 2 + 4λ – 3 + 16λ + 8 + 5 = 0

24λ + 12 = 0 λ = ½

So, the coordinates of the point in the plane are

2(-1/2) + 1, -2(-1/2) + 3/2, 4(-1/2) + 2 = 0, 5/2, 0

Thus, the foot of the perpendicular is (0, 5/2, 0) and the required length

$=\sqrt{(1-0)^{2}+\left(\frac{3}{2}-\frac{5}{2}\right)^{2}+(2-0)^{2}}$

$=\sqrt{1+1+4}=\sqrt{6}$ units