Question

Prove that if a plane has the intercepts a , b , c and is at a distance of P units from the origin, then

Answer 1

The intercepts of the plane are a,b,c. The distance of the intercept from the origin is p.

The equation of the plane having intercepts a,b,c with x,y,z respectively is given by,

x a + y b + z c =1(1)

The formula for the distance of the plane Ax+By+Cz+d=0 from point ( x 1 , y 1 , z 1 ) is given by,

p=| A x 1 +B y 1 +C z 1 +d A 2 + B 2 + C 2 |(2)

Here,

A= 1 a B= 1 b C= 1 c

Substitute the values of A, B, C and ( x 1 , y 1 , z 1 ) as ( 0,0,0 ) in equation (2).

p=| 1 a ( 0 )+ 1 b ( 0 )+ 1 c ( 0 )+( −1 ) ( 1 a ) 2 + ( 1 b ) 2 + ( 1 c ) 2 | p= 1 1 a 2 + 1 b 2 + 1 c 2

Square on both the sides,

p 2 = 1 1 a 2 + 1 b 2 + 1 c 2 1 p 2 = 1 a 2 + 1 b 2 + 1 c 2

Hence, the plane has an intercept of a,b,c and the distance of p units from the origin is 1 p 2 = 1 a 2 + 1 b 2 + 1 c 2 .