The line passing through the points $(5,1, a)$ and $(3, b, 1)$ crosses the $y z-$ plane at the point…

The line passing through the points $(5,1, a)$ and $(3, b, 1)$ crosses the $y z-$ plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$. Then
  1. $a=2, b=8$
  2. $a=4, b=6$
  3. $a=6, b=4$
  4. $a=8, b=2$

Solution

Equation of line passing through $(5,1, a)$ and $(3, b, 1)$ is $ \frac{x-5}{2}=\frac{y-1}{1-b}=\frac{z-a}{a-1}=\lambda $ If line crosses $y z-$ plane i.e., $x=0$ $ \begin{aligned} & x=2 \lambda+5=0 \\ & \Rightarrow \lambda=-5 / 2, \end{aligned} $ Since, $y=\lambda(1-b)+1=\frac{17}{2}$ $ \begin{aligned} & -\frac{5}{2}(1-b)+1=\frac{17}{2} \\ & b=4 \end{aligned} $ Also, $z=\lambda(a-1)+a=-\frac{13}{2}$ $ \begin{aligned} & -\frac{5}{2}(a-1)+a=-\frac{13}{2} \\ & \Rightarrow a=6 \end{aligned} $

Asked in: JEE Main 2008

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