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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.
Solution
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.`
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