Question

Find the distance of the point (2,2-1) from the plane x+2y-z=1 measured parallel to the line x+3/2=

Answer 1

fine the distance of the point ,(2.2-1) from the plane× 26 y/-z=7

Answer 2

GIVEN :

◆The point (2,2-1) , from where distance is to be calculated.

◆The plane x+2y-z=1 measured parallel to the line x+3/2.

SOLUTION :

◆Equation of line from point (2,2,-1 ) to the line to be measured {x+3/2} is

(x - 2) + 3/2 = x - 1 /2 = l

x = l + 1/2 ; y =0 ; z= 0

◆Substituting this values of x in equation of plane ,

x + 2y - z - 1 = l + (1/2 ) + 0 -1 =0

l - 1/2 =0

l = 1/2

x = l + 1/2 = 1; y = 0; z =0

ANSWER :

◆Distance between point (2,2,-1) from the plane parallel to the line is

d = [( x - x' )^2 + (y - y')^2 +(z - z')^2]^(1/2)

d^2 =( 1 - 2)^2 + (0-2)^2 +(0--1)^2

=1+ 4 +1 = 6cm.

d =√6 cm.