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Find the equation of the plane containing the lines 2x-y+z-3=0,3x+y+z=5 and a t a distance of `1/sqrt6` from the point (2,1,-1). |
Answer» Correct Answer - `2x-y+z-3=0;62x+29y+19z-105=0` Equation of plane containing the lines `2x-y+z-3=0" and "3x+y+z=5` is `(2x-y+z-3)+lambda(3x+y+z-5)=0` `implies(2+3lambda)x+(lambda-1)y+(lambda+1)z-3-5lambda=0` Since, distance of plane from (2,1,-1) to above plane is `1sqrt(6).` `:.|(6lambda+4+lambda-1-lambda-1-3-5lambda)/(sqrt((3lambda+2)^(2)+(lambda-1)^(2)+(lambda+1)^(2)))|=(1)/(sqrt(6))` `implies" "6(lambda-1)^(2)=11lambda^(2)+12lambda+6` `implies" "lambda=0,-(24)/(5)` `:.` Equations of planes are `2x-y+z-3=0" and "62x+29y+19z-105=01` |
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