Find the vector equation of the plane which contains the line of intersection of the planes `vecr (hati+2hatj+3hatk)-4=0` and `vec r (2hati+hatj-hatk)+5=0` which is perpendicular to the plane.`vecr(5hati+3hatj-6hatk)+8=0`
Solution
The equations of the given planes are
`vecr (hati+2hatj+3hatk)-4=0 ........(1)`
`vec r (2hati+hatj-hatk)+5=0 .........(2)`
The equation of the plane passing through the intersection of the planes (1) and (2) is
`[vecr (hati+2hatj+3hatk)-4]+lambda[ vec r (2hati+hatj-hatk)+5]=0`
`vecr[(1+2lambda)hati+(2+lambda)hatj+(3-lambda)hatk]=4-5lambda.............(3)`
Given, the plane (3) is perpendicular to the plane `vecr(5 hati+3hatj-6hatk)+8=0`
`(1+2lambda)xx5+(2+lambda)xx3+(3-lambda)xx(-6)=0`
`19lambda-7=0`
`lambda=7/19`
Putting `lambda=7/19 ` in (3), we get
`vec r[(1+14/19)hati+(2+7/19)hatj+(3-7/19)hatk]=4-35/19`
`vecr(33/19hati+45/19hatj+50/19hatk)=41/19`
`vecr(33hati+45hatj+50hatk)=41`
Thus, the equation of the required plane is
`vecr(33hati+45hatj+50hatk).=41`
