Question 310 of 521MediumMCQ
Equation of a plane at a distance 212 from the origin, which contains the line of intersection of the planes x−y−z−1=0 and 2x+y−3z+4=0, is Solutions · 1
Official solution
Required equation of plane
\[{P_1} + \lambda {P_2} = 0\]
\[(x - y - z - 1) + \lambda (2x + y - 3z + 4) = 0\]
Given that its dist. From origin is \[{2 \over {\sqrt {21} }}\]
Thus, \[{{|4\lambda - 1|} \over {\sqrt {{{(2\lambda + 1)}^2} + {{(\lambda - 1)}^2} + {{( - 3\lambda - 1)}^2}} }} = {{\sqrt 2 } \over {\sqrt {21} }}\]
\[ \Rightarrow 21{(4\lambda - 1)^2} = 2(14{\lambda ^2} + 8\lambda + 3)\]
\[ \Rightarrow 336{\lambda ^2} - 168\lambda + 21 = 28{\lambda ^2} + 16\lambda + 6\]
\[ \Rightarrow 308{\lambda ^2} - 184\lambda + 15 = 0\]
\[ \Rightarrow 308{\lambda ^2} - 154\lambda - 30\lambda + 15 = 0\]
\[ \Rightarrow (2\lambda - 1)(154\lambda - 15) = 0\]
\[ \Rightarrow \lambda = {1 \over 2}\] or \[{{15} \over {154}}\]
for \[\lambda = {1 \over 2}\] reqd. plane is 4x−y−5z+2=0 Answer the question to unlock