question_answer

An equation of the plane through the points (1, 0, 0) and (0, 2, 0) and at a distance\[\frac{6}{7}\]units from the origin is

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,                 or