Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati+hatj+hatk)=1 and vecr.(2hati+3hatj-hatk)+4=0` and parallel to x-axis.

Explanation

Converting the given equations of the planes in Cartesian form, we get
`vecr.(hati+hatj+hatk)=1 rArr (xhati+yhatj+zhatk).(hati+hatj+hatk)=1`
`rArr x+y+z-1=0`….……….(i)
`vecr.(2hati+3hatj-hatk)+4=0 rArr (xhati+yhatj+zhatk).(2hati+3hatj-hatk)+4=0`
`rArr 2x+3y-z+4=0`,...........(ii)
Now, the equation of a plane passing through the intersection of the planes (i) and (ii) is given by
`(x+y+z-1)+lambda(2x+3y-z+4)=0` for some real value of `lambda`
`rArr (1+2lambda)x+(1+3lambda)y(1-lambda)z+(-1+4lambda)=0`...........(iii)
D.r.'s of the normal to the plane are `(1+2lambda),(1+3lambda),(1-lambda)`.
D'r'.s of the x-axis are 1,0,0.
Plane (iii) is parallel to the x-axis means that the normal to this plane is perpendicular to the x-axis.
`therefore 1 xx (1 + 2lambda)+0 xx (1 + 3lambda)+0 xx (1-lambda)=0 rArr 1+2lambda=0 rArr lambda=-1/2`.
Putting `lambda=-1/2` in (iii), we get the required equation of the plane as
`(1-1)x+(1-3/2)y+(1+1/2)z+(-1-2)=0`
`rArr -y/2+(3z)/2 -3=0 rArr y-3z+6=0`.
Hence, the required equation of the plane is `y-3z+6=0`.