Let the plane \[P:\overrightarrow r \,.\,\overrightarrow a = d\] contain the line of intersection of two planes \[\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 6\] and \[\overrightarrow r \,.\,\left( { - 6\widehat i + 5\widehat j - \widehat k} \right) = 7\]. If the plane P passes through the point \[\left( {2,3,{1 \over 2}} \right)\], then the value of \[{{|13\overrightarrow a {|^2}} \over {{d^2}}}\] is equal to :
Solutions · 1
Official solution
P1:x+3y−z=6P2:−6x+5y−z=7
Family of planes passing through line of intersection of P1 and P2 is given by \[x(1 - 6\lambda ) + y(3 + 5\lambda ) + z( - 1 - \lambda ) - (6 + 7\lambda ) = 0\]
It passes through \[\left( {2,3,{1 \over 2}} \right)\]
So, \[2(1 - 6\lambda ) + 3(3 + 5\lambda ) + {1 \over 2}( - 1 - \lambda ) - (6 + 7\lambda ) = 0\]
\[ \Rightarrow 2 - 12\lambda + 9 + 15\lambda - {1 \over 2} - {\lambda \over 2} - 6 - 7\lambda = 0\]
\[ \Rightarrow {9 \over 2} - {{9\lambda } \over 2} = 0 \Rightarrow \lambda = 1\]
Required plane is
−5x+8y−2z−13=0
Or \[\overrightarrow r .\,( - 5\widehat i + 8\widehat j - 2\widehat k) = 13\]
\[{{|13\overrightarrow a {|^2}} \over {|d{|^2}}} = {{{{13}^2}} \over {{{(13)}^2}}}.\,|\overrightarrow a {|^2} = 93\]