Find the equation of the plane containing the line `(y)/(b)+(z)/(c)=1,x=0`, and parallel to the line `(x)/(a)-(z)/(c)=1,y=0`.

Answer

`(x)/(a)-(y)/(b)-(z)/(c )+1=0`

Explanation

The equation of the first line may be written as
`(y)/(b)-(1)/(2)=(1)/(2)-(z)/(c),x=0`
or `(x)/(0)=(y-(1)/(2)b)/(b)=(z-(1)/(2)c)/(-c)" ".....(i)`
Similarly, the equation of the second line may be written as
`(x-(1)/(2)a)/(a)=(y)/(0)=(z+(1)/(2)c)/(c)" ".....(ii)`
The equation of any plane passing through line (i) is
`A(x)+B(y-(1)/(2)b)+C(z-(1)/(2)c)=0" "....(iii)`
where `A.0+B.b-C.c=0" ".....(iv)`
Now plane (iii) will be parallel to line (ii) if
`A.a+B.b-C.c=0" ".....(v)`
Solving (iv) adn (v), we have `(A)/(bc)=(B)/(-ca)=(C)/(-ab)`
Putting these values of A, B and C in (iii), the equation of the required plane is
`bcx-ca(y-(1)/(2)b)=ab(z-(1)/(2)c)=0`
or `(x)/(a)-(y)/(b)-(z)/(c)+1=0`