The equation of the line, passing through $(1,2,3)$ and parallel to planes $x-y+2 z=5$ and $3 x+y+z=6$, is
The equation of the line, passing through $(1,2,3)$ and parallel to planes $x-y+2 z=5$ and $3 x+y+z=6$, is
- $\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
- $\frac{x-1}{-3}=\frac{y-2}{-5}=\frac{z-3}{4}$
- $\frac{x-1}{4}=\frac{y-2}{5}=\frac{z-3}{3}$
- $\frac{x-1}{5}=\frac{y-2}{7}=\frac{z-3}{1}$
Solution
Required equation of line is
$\begin{aligned}
& \frac{x-1}{\left|\begin{array}{cc}
-1 & 2 \\
1 & 1
\end{array}\right|}=\frac{y-2}{-\left|\begin{array}{ll}
1 & 2 \\
3 & 1
\end{array}\right|}=\frac{\mathrm{z}-3}{\left|\begin{array}{cc}
1 & -1 \\
3 & 1
\end{array}\right|} \\
\therefore \quad & \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}
\end{aligned}$
Asked in: MHT CET 2023 (12 May Shift 2)
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