Question

The equation of the plane, which is equidistant from lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-2}{3}$ and $\frac{x-2}{2}=\frac{y-2}{1}=\frac{z-1}{2}$ is

• A. $3x+2y-3z=10$
• B. $x+7y-5z=9$
• C. $x+4y-3z=5$
• D. $x+19y-5z=11$

Answer 1

The correct option is C $x+4y-3z=5$

Normal vector of required plane is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 2& 1& 2\end{array}\right|=\hat{i}+4\hat{j}-3\hat{k}$
Let required plane be $P:x+4y-3z=c$
Given that plane is equidistant from the lines.
So, distances of plane $P$ from points $(1,2,2)$ and $(2,2,1)$ are equal
$\Rightarrow \frac{|1+4\times 2-3\times 2-c|}{\sqrt{1^2+4^2+(-3{)}^2}}=\pm \frac{|2+4\times 2-3\times 1-c|}{\sqrt{1^2+4^2+(-3{)}^2}}\Rightarrow 3-c=-(7-c)\Rightarrow c=5$
Hence, required plane is $x+4y-3z=5$