A) (-1, 3, 2)
B) (-1, -3, 2)
C) (2, 1, 3)
D) (2, 3, 2)
Correct Answer: A
Solution :
[a] The intersection of given plane is \[x-y+2z-1+\lambda (x+y-z-3)=0\] \[\Rightarrow x(1+\lambda )+y(\lambda -1)+z(2-\lambda )-3\lambda -1=0\] DR?s of normal to the above plane is \[(1+\lambda ,\lambda -1,2-\lambda )\] By taking option (a) \[-1(1+\lambda )+3(\lambda -1)+2(2-\lambda )=0\] \[\Rightarrow -1-\lambda +3\lambda -3+4-2\lambda =0\] \[\Rightarrow 0=0\] Which is true.You need to login to perform this action.
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