A) \[0<a<\frac{1}{3}\]
B) \[-1<a<0\]
C) \[a<-1\]or \[a<\frac{1}{3}\]
D) \[a=0\]
Correct Answer: D
Solution :
Given equation of plane is \[\text{3x}+\text{4y}-\text{12z}+\text{13}=0\] (1, a, 1) and (-3,0,a) satisfy the equation of plane. \[\therefore \]We have\[\text{3}+\text{4(a)}-\text{12}+\text{13}=0\]and\[\text{3(-3)}-\text{12(a)4}-\text{13}=0\]\[\Rightarrow \]\[4+4a=0\]and\[4-12a=0\]\[\Rightarrow \]\[a=-1\]and\[a=\frac{1}{3}\]Since, (1, a, 1) and (- 3, 0, a) lie on the opposite sides of the plane \[\therefore \]\[a=0\]
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