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Question
Find the equation of the plane which contains the line of intersection of the planes
`vecr.(hati-2hatj+3hatk)-4=0" and"`
`vecr.(-2hati+hatj+hatk)+5=0`
and whose intercept on x-axis is equal to that of on y-axis.
Solution
`vecr.(hati-2hatj+3hatk)-4=0`
`vecr.(-2hati+hatj+hatk)+5=0`
`vecr.(hati-2hatj+3hatk)+lambda{vecr.(-2hati+hatj+hatk)}-4+5lambda = 0`
`=>vecr.[(1-2lambda)hati+(-2+lambda)hatj+(3+lambda)hatk]-4+5lambda=0`
`=>(1-2lambda)x+(-2+lambda)y+(3+lambda)z=-5lambda+4`
`=>x/((-5lambda+4)/(1-2lambda))+y/((-5lambda+4)/(-2+lambda))+z/((-5lambda+4)/(3+lambda))=1`
`:.(-5lambda+4)/(1-2lambda) =(-5lambda+4)/(-2+lambda)`
⇒ 1- 2λ = -2 + λ
⇒ -3λ = -3
⇒ λ = 1
∴ Equation of the required plane
- x - y + 4z = -1
x + y - 4z - 1 = 0
Vector eqn of the required Plane
`=>vecr.(hati+hatj-4hatk)-1=0`
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