The Cartesian equation of the plane passing through the point $(0,7,-7)$ and containing the line…
The Cartesian equation of the plane passing through the point $(0,7,-7)$ and containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ is
- $2 x+y-z=14$
- $x+y+z=0$
- $x+2 y+z=7$
- $2 x+y+z=0$
Solution
The plane passes through the point $(0,7,-7)$ contains the line $\frac{\mathrm{x}+1}{-3}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}+2}{1}$.
$\therefore$ Required equation of the plane is
$\begin{aligned}
& \left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-1-0 & 3-7 & -2+7 \\
-3 & 2 & 1
\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-1 & -4 & 5 \\
-3 & 2 & 1
\end{array}\right|=0 \\
& \therefore \hat{\mathrm{i}}(-4-10)-\hat{\mathrm{j}}(-1+15)+\hat{\mathrm{k}}(-2-12)=0
\end{aligned}$
Hence Cartesian equation of the plane is $x+y+z=0$
Asked in: MHT CET 2021 (20 Sep Shift 1)
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