Question
The line passing through the points $$\left( {5,\,1,\,a} \right)$$ and $$\left( {3,\,b,\,1} \right)$$ crosses the $$yz$$ -plane at the point $$\left( {0,\,\frac{{17}}{2},\,\frac{{ - 13}}{2}} \right).$$ Then :
A.
$$a = 2,\,b = 8$$
B.
$$a = 4,\,b = 6$$
C.
$$a = 6,\,b = 4$$
D.
$$a = 8,\,b = 2$$
Answer :
$$a = 6,\,b = 4$$
Solution :
Equation of line through $$\left( {5,\,1,\,a} \right)$$ and $$\left( {3,\,b,\,1} \right)$$ is $$\frac{{x - 5}}{{ - 2}} = \frac{{y - 1}}{{b - 1}} = \frac{{z - a}}{{1 - a}} = \lambda $$
$$\therefore $$ Any point on this line is a $$\left[ { - 2\lambda + 5,\,\left( {b - 1} \right)\lambda + 1,\,\left( {1 - a} \right)\lambda + a} \right]$$
It crosses $$yz$$ plane where $$ - 2\lambda + 5 = 0 \Rightarrow \lambda = \frac{5}{2}$$
$$\eqalign{
& \therefore \,\left( {0,\,\left( {b - 1} \right)\frac{5}{2} + 1,\,\left( {1 - a} \right)\frac{5}{2} + a} \right) = \left( {0,\,\frac{{17}}{2},\,\frac{{ - 13}}{2}} \right) \cr
& \Rightarrow \left( {b - 1} \right)\frac{5}{2} + 1 = \frac{{17}}{2}{\text{ and }}\left( {1 - a} \right)\frac{5}{2} + a = - \frac{{13}}{2} \cr
& \Rightarrow b = 4{\text{ and }}a = 6 \cr} $$