InterviewSolution
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InterviewSolution
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In `R^(3)`, consider the planes `P_(1):y=0` and `P_(2),x+z=1.` Let `P_(3)` be a plane, different from `P_(1)` and `P_(2)` which passes through the intersection of `P_(1)` and `P_(2)`, If the distance of the point (0,1,0) from `P_(3)` is 1 and the distance of a point `(alpha,beta,gamma)` from `P_(3)` is 2, then which of the following relation(s) is/are true?A. `2alpha+beta+2gamma+2=0`B. `2alpha-beta+2gamma+4=0`C. `2alpha+beta-2gamma-10=0`D. `2alpha-beta+2gamma-8=8` |
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Answer» Correct Answer - B::D The plane passing through the intersection of planes `P_(1):y=0` and `P_(2):x+z=1` is `P_(3):(x+z-1)+lamday=0` or `x+lamday+z-1=0` ………..i It is given that the distances of the point `(0,1,0)` and `(alpha, beta, gamma)` from i are 1 and 2 respectively. `:.|(0+lamda+0-1)/(sqrt(1+lamda^(2)+1))|=1` and `|(alpha+lamda beta+gamma-1)/(sqrt(1+lamda^(2)+1))|=2` `implies(|lamda-1|)/(sqrt(lamda^(2)+2))=1` and `(|alpha+lamda beta+gamma-1|)/(sqrt(lamda^(2)+2))=2` `implies(lamda-1)^(2)=lamda^(2)+2` and `|alpha+lamda beta+gamma-1|=2sqrt(lamda^(2)+2)` `implieslamda=-1/2` and `|alpha+lamda beta+gamma-1|-2sqrt(lamda^(2)+2)` `implies|2alpha-beta+2gamma-2|=6` `implies2alpha-beta+2gamma=+-2alpha-beta+2gamma-8=0` or `2alpha-beta+2gamma+4=0` |
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