The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0…
The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is
- $2 x+y+z=0$
- $x+y+z=0$
- $x+2 y-3 z=35$
- $x+3 y+z=14$
Solution
Let $a, b, c$ be the direction cosines of the required plane.
It contains the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and passes through the point $(0,7,-7)$
$\begin{aligned}
& \therefore \mathrm{a}(\mathrm{x}+1)+\mathrm{b}(\mathrm{y}-3)+\mathrm{c}(\mathrm{z}+2)=0 \quad \ldots(1) \\
& \therefore \mathrm{a}(0+1)+\mathrm{b}(7-3)+\mathrm{c}(-7+2)=0 \Rightarrow \mathrm{a}+4 \mathrm{~b}-5 \mathrm{c}=0 .
\end{aligned}$
$\text { Also }-3 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=0$
From (2) and (3), we write
$\frac{\mathrm{a}}{\left|\begin{array}{cc}
4 & -5 \\
2 & 1
\end{array}\right|}=\frac{-b}{\left|\begin{array}{cc}
1 & -5 \\
-3 & 1
\end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc}
1 & 4 \\
-3 & 2
\end{array}\right|}$
$\therefore \frac{\mathrm{a}}{14}=\frac{\mathrm{b}}{14}=\frac{\mathrm{c}}{14} \Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}=1$
Hence eq. (1) becomes
$x+1+y-3+z+2=0 \Rightarrow x+y+z=0$
Asked in: MHT CET 2021 (21 Sep Shift 2)
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