Question

Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.

Answer 1

The given equation of the plane is 2x + y - 2z = 3

Dividng both sides by 3, we get

`(2x)/3 + y/3 +(-2z)/3 = 3/3 `

`⇒ x/((3/2)) + y/3 + z/(((-3)/2)) = 1      ...      (1) `

We know that the equation of the plane whose intercepts on the coordianate axes are  

`x/a + y/b +z/c = 1`  .......... (2) 

Comparing (1) and (2), we get

`a = 3/2 ;  b =3 ; c = (-3)/2`

Finding the direction cosines of the normalThe given equation of the plane is 

2x + y - 2z = 8

`⇒ (x  hat(i) + y   hat(j) + z hat(k) ). (2  hat(i) + hat(j) - 2 hat(k)) = 8`

`⇒ vec r . (2  hat(i) + hat(j) - 2 hat(k)) = 8 , ` which is the vector equation of the plane.(Because the vector equation of the plane is `vec r . vec n = vec a . vec n ,`  where the normal to the plane  `vec n = 2    vec i + vec j -2 hat(k). )`

`|vec n| = sqrt (4+1+4) =3`

So, the unit vector perpendicular to` vec n =(vec n)/(|vec n |) = ( 2     hat(i)+hat(j)-2hat(k))/3 = 2/3 hat(i) +1/3 hat(j) -2/3 hat(k) `

So, the direction cosines of the normal to the plane are, ` 2/3 , 1/3  ,(-2)/3.`