Question 431 of 521MediumMCQ
The line passing through the points (5,1,a) and (3,b,1) crosses the yz-plane at the point (0,217,−213). Then Solutions · 1
Official solution
Equation of line through \[\left( {5,1,a} \right)\] and
\[\left( {3,b,1} \right)\] is \[{{x - 5} \over { - 2}} = {{y - 1} \over {b - 1}} = {{z - a} \over {1 - a}} = \lambda \]
\[\therefore\] Any point on this line is a
\[\left[ { - 2\lambda + 5,\left( {b - 1} \right)\lambda + 1,\left( {1 - \alpha } \right)\lambda + a} \right]\]
It crosses yz plane where \[{ - 2\lambda + 5 = 0}\]
\[\lambda = {5 \over 2}\]
\[\therefore\] \[\left( {0,\left( {b - 1} \right){5 \over 2} + 1,\left( {1 - a} \right){5 \over 2} + a} \right) = \left( {0,{{17} \over 2},{{ - 17} \over 2}} \right)\]
\[ \Rightarrow \left( {b - 1} \right){5 \over 2} + 1 = {{17} \over 2}\]
and \[\left( {1 - a} \right){5 \over 2} + a = - {{13} \over 2}\]
\[ \Rightarrow b = 4\] and a=6 Answer the question to unlock