find cartesian eqn & vector eqn of the planes passing thru the intersection of the planes r(2i+6j)+12=0 and r(3i-j+4k)=0 which is at unit distance from origin
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Equation of given planes are
r.(2i+6j)+12=0 ---(1) and r.(3i-j+4k)=0---(2)
Now, Let the Equation of plane passing through the intersection of these planes be
r.(2i+6j)+12+λ[r.(3i-j+4k)]=0
=> r.{(2+3λ)i+(6-λ)j+4λk}+12=0 ---(3)
Distance of this plane from the origin ,
12/√(2+3λ)2+(6-λ)2+(4λ)2=1
Squaring the Given Expression
144= 4+9λ2+12λ+36+λ2-12λ+16λ2
=> 144-40= 26λ2
=> λ2=104/26
=> λ2=4
=> λ=+- 2
Putting the values in Equation (3), we get
r.{(8)i+(4)j+8k}+12=0 and r.{(-4)i+(8)j-8k}+12=0
Equation of given planes are
r.(2i+6j)+12=0 ---(1) and r.(3i-j+4k)=0---(2)
Now, Let the Equation of plane passing through the intersection of these planes be
r.(2i+6j)+12+λ[r.(3i-j+4k)]=0
=> r.{(2+3λ)i+(6-λ)j+4λk}+12=0 ---(3)
Distance of this plane from the origin ,
12/√(2+3λ)2+(6-λ)2+(4λ)2=1
Squaring the Given Expression
144= 4+9λ2+12λ+36+λ2-12λ+16λ2
=> 144-40= 26λ2
=> λ2=104/26
=> λ2=4
=> λ=+- 2
Putting the values in Equation (3), we get
r.{(8)i+(4)j+8k}+12=0 and r.{(-4)i+(8)j-8k}+12=0
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