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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
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Verified Answer
The correct answer is:
$a=6, b=4$
$a=6, b=4$
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, $z=\lambda(a-1)+a=-\frac{13}{2}$
$$
\begin{aligned}
& -\frac{5}{2}(a-1)+a=-\frac{13}{2} \\
& \Rightarrow a=6
\end{aligned}
$$
$$
\frac{x-5}{2}=\frac{y-1}{1-b}=\frac{z-a}{a-1}=\lambda
$$
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, $z=\lambda(a-1)+a=-\frac{13}{2}$
$$
\begin{aligned}
& -\frac{5}{2}(a-1)+a=-\frac{13}{2} \\
& \Rightarrow a=6
\end{aligned}
$$
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