A line with direction cosines proportional to 2,1,2 meets each of the lines x=y+a=z and x+a=2y=2z. The co-ordinates of each of the points of intersection are given by
Solutions · 1
Official solution
Let a point on the line
x=y+a=z is \[\left( {\lambda ,\lambda - a,\lambda } \right)\]
and a point on the line
x+a=2y=2z is \[\left( {\mu - a,{\mu \over 2},{\mu \over 2}} \right),\]
then direction ratio of the line joining these points are
\[\lambda - \mu + a,\,\,\lambda - a - {\mu \over 2},\,\,\lambda - {\mu \over 2}\]
If it represents the required line, then
\[{{\lambda - \mu + a} \over 2}\]
\[ = {{\lambda - a - {\mu \over 2}} \over 1}\]
\[ = {{\lambda - {\mu \over 2}} \over 2}\]
on solving we get \[\lambda = 3a,\,\mu = 2a\]
\[\therefore\] The required points of intersection are
\[\left( {3a,3a - a,3a} \right)\] and \[\left( {2a - a,{{2a} \over 2},{{2a} \over 2}} \right)\]
or \[\left( {3a,2a,3a} \right)\] and \[\left( {a,a,a} \right)\]