Question

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

Answer 1

This is the required answer

Answer 2

Given :

A point ( 1, 3/2 ,2 )

A plane 2x – 2y + 4z + 5 = 0

To Find :

Length & foot of perpendicular from

point ( 1, 3/2 ,2 ) .

Solution:

•Foot of perpendicular passes from point ( 1, 3/2 ,2 ) .

•Also , direction ratio of perpendicular to plane will be same as that of plane i.e. 2 , -2 , 4 or

1 , -1 ,2.

•Equation of perpendicular is

(x-1)/1 = (y-3/2)/-1 =( z-2)/2

•Let a general point on line

(x-1)/1 = (y-3/2)/-1 =( z-2)/2 = k _(say)

x = k + 1

y = 3/2 - k

z = 2k + 2

•If this point lies on plane then , it will satisfy the equation of plane.

2x – 2y + 4z + 5 = 0

x – y + 2z + 5/2 = 0

(k+1) -(3/2 - k ) +2(2k + 2 ) + 5/2 = 0

6k + 12/2 = 0

6k + 6 = 0

k = -1

•So , Foot of perpendicular will be

x = k + 1 = -1+1 = 0

y = 3/2 - k = 3/2-(-1) = 5/2

z = 2k + 2 = -2 + 2 = 0

•Foot of perpendicular will be (0,5/2,0)

•By distance formula length of perpendicular is

√[(X2-X1)² + (Y2-Y1)² + (Z2-Z1)²]

√[(0-1)² + (5/2-3/2)² + (0-2)²]

√[(1)² + (1)² +(-2)²]

√[1+1+4]

√6 units