Find the vector of the plane passing through the intersection of the planes `vecr.(2hati+2hatj-3hatk)=7, vecr.(2hati+5hatj+3hatk)=9` and the point (2,1,3)`.
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Equation of given planes are
`vecr.(2hati+2hatj-3hatk)=7` and `vecr.(2hati+5hatj+3hatk) = 9`
The equation of these planes can be written as
` v ecr.(2hati+2hatj-3hatk) – 7 = 0 “………”(1)`
and `vecr.(2hati+5hatj+3hatk) – 9 = 0 “…..”(2)`
Equation of a plane through the intersection of plane (1) and (2) is,
`[vecr.(2hati+2hatj-3hatk)-7]+lambda[vecr.(2hati+5hatj+3hatk)-9]=0`
`rArr vecr.[(2hati+2hatj-3hatk)+lambda(2hati+5hatj+3hatk)]=9lambda+7`
`rArr [vecr.(2+2lambda)hati+(2+5lambda)hatj+(3lambda-3)hatk]=9lambda+7″…….”(3)`
This intersecting plane passes through the point (2,1,3), whose position vector is `vecr.(2hati+hatj+3hatk)`. Put this value of `vecr` in eqaution (3),
`(2hati+hatj+ 3hatk) .[(2+2lambda) hati+(2+5lambda)hatj+(3lambda-3)hatk] = 6 lambda + 7`
`rArr 2(2+2lambda)+(2+5lambda)+3(3lambda-3)=9lambda+7`
`rArr (4+4lambda)+(2+5lambda)+(9lambda-9)=9lambda+7`
`rArr -3+18lambda=9lambda+7`
`rArr 9lambda = 10 lambda = 10/9`
`:.` Equation of plane
`[vecr.(2hati+2hatj-3hatk)-7]+10/9[vecr.(2hati+5hatj+3hatk)-9]=0`
`rArr vecr .(18hati+18hatj-27hatk+20hati+50hatj+30hatk)-63-90=0`
`rArr vecr.(38hati+68hatj+3hatk) = 153`
which is the required equation of the plane.