Q. 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
KEAMKEAM 2011
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,
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