If the line, $\frac{x-3}{2}=\frac{y+2}{1}=\frac{z+4}{3}$ lies in the plane, $\ell x+m y-z=9$, then…
If the line, $\frac{x-3}{2}=\frac{y+2}{1}=\frac{z+4}{3}$ lies in the plane, $\ell x+m y-z=9$, then $\ell^2+m^2$ is equal to
- $\frac{124}{49}$
- $\frac{123}{49}$
- $\frac{121}{49}$
- $\frac{122}{49}$
Solution
Line is perpendicular to normal to plane
$\begin{aligned}
& \Rightarrow(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(l \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\
& 2 l+\mathrm{m}-3=0...(i)
\end{aligned}$
$(3,-2,-4)$ lies on the plane
$l x+\mathrm{m} y-\mathrm{z}=9$
$\begin{aligned}
& \therefore \quad 3 l-2 \mathrm{~m}+4=9 \\
& \Rightarrow 3 l-2 \mathrm{~m}=5...(ii)
\end{aligned}$
Solving (i) and (ii), we get
$\begin{aligned}
& l=\frac{11}{7}, \mathrm{~m}=\frac{-1}{7} \\
& l^2+\mathrm{m}^2\left(\frac{11}{7}\right)^2+\left(\frac{-1}{7}\right)^2=\frac{122}{49}
\end{aligned}$
Asked in: MHT CET 2024 (10 May Shift 1)
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