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Question

# Find the vector equation of the plane passing through the intersection of the planes →r⋅(^i+^j+^k)=6 and →r⋅(2^i+3^j+4^k)=−5 and the point (1,1,1).

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Solution

## Here, →n1=^i+^j+^k and →n2=^2i+3^j+4^k d1=6 and d2=−5 Hence, using the relation →r.(→n1+λ→n2)=d1+λd2 , we get →r⋅[(^i+^j+^k)+λ(^2i+3^j+4^k)]=6−5λ ⇒→r⋅[(1+2λ)^i+(1+3λ)^j+(1+4λ)^k]=6−5λ ⋯(1) where, λ is some real number. Taking →r=x^i+y^j+z^k, we get (x^i+y^j+z^k).[(1+2λ)^i+(1+3λ)^j+(1+4λ)^k]=6−5λ ⇒(1+2λ)x+(1+3λ)y+(1+4λ)z=6−5λ ⇒(x+y+z−6)+λ(2x+3y+4z+5)=0 ⋯(2) Given that the plane passes through the point (1,1,1), it must satisfy equation(2), i.e., (1+1+1−6)+λ(2+3+4+5)=0 ⇒−3+λ(14)=0⇒λ=314 Putting the value of λ in (1), we get ⇒→r⋅[(1+37)^i+(1+914)^j+(1+67)^k]=6−1514 ⇒→r⋅(107^i+2314^j+137^k)=6914 ⇒→r⋅(20^i+23^j+26^k)=69 Which is the required vector equation of the plane.

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