A) \[6x+3y+z-6=0\]
B) \[6x+3y+2z-6=0\]
C) \[6x+3y+z+6=0\]
D) \[6x+3y+2z+6=0\]
E) \[6x+2y+3z+6=0\]
Correct Answer: B
Solution :
The equation of plane passing through (1, 0, 0) is \[a(x-1)+b(y-0)+c(z-0)=0\] ...(i) Since, plane also passing through (0, 2, 0). \[\Rightarrow \] \[a(0-1)+b(2-0)+c(0-0)=0\] \[\Rightarrow \] \[-a+2b=0\] \[\Rightarrow \] \[a=2b\] ...(ii) Given, distance from origin to plane (i) =6/7 \[\frac{|a(0-1)+b(0-0)+c(0-0)|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\frac{6}{7}\] \[\Rightarrow \] \[\left| \frac{-a+0+0}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|=\frac{6}{7}\] \[\Rightarrow \] \[\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\frac{6}{7}\] \[\Rightarrow \] \[\frac{2b}{\sqrt{4{{b}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\frac{6}{7}\] [from Eq.(ii)] \[\Rightarrow \] \[14b=6\sqrt{5{{b}^{2}}+{{c}^{2}}}\] Squaring on both sides; \[\Rightarrow \] \[196{{b}^{2}}=36(5{{b}^{2}}+{{c}^{2}})\] \[\Rightarrow \] \[196{{b}^{2}}=180{{b}^{2}}+36{{c}^{2}}\] \[\Rightarrow \] \[16{{b}^{2}}=36{{c}^{2}}\] \[\Rightarrow \] \[4b=6c\] ...(iii) From Eq. (ii) and Eq. (iii), So, required equation of plane is, orYou need to login to perform this action.
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