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The line joining the points $(3,5,-7)$ and $(-2,1,8)$ meets the $y z$-plane at point
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Verified Answer
The correct answer is:
$\left(0, \frac{13}{5}, 2\right)$
Line joining the points $(3,5,-7)$ and $(-2,1,8)$ is, $\frac{x-3}{(-2)-(3)}=\frac{y-5}{(1)-(5)}=\frac{z-(-7)}{8-(-7)}$
$\begin{gathered}
\frac{x-3}{-5}=\frac{y-5}{-4}=\frac{z+7}{15}=K \\
\therefore x=-5 K+3, y=-4 K+5, \quad z=15 K-7
\end{gathered}$
L. Line (i) meets the $y z$-plane
$\therefore \quad-5 K+3=0 \Rightarrow K=3 / 5$
Put the value of $K$ in $x, y, z$
So the required point is $(0,13 / 5,2)$.
$\begin{gathered}
\frac{x-3}{-5}=\frac{y-5}{-4}=\frac{z+7}{15}=K \\
\therefore x=-5 K+3, y=-4 K+5, \quad z=15 K-7
\end{gathered}$
L. Line (i) meets the $y z$-plane
$\therefore \quad-5 K+3=0 \Rightarrow K=3 / 5$
Put the value of $K$ in $x, y, z$
So the required point is $(0,13 / 5,2)$.
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