Q80 / 150 · MHT CET 2024 (10 May Shift 1) · Mathematics

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

  • A
    $\frac{124}{49}$
  • B
    $\frac{123}{49}$
  • C
    $\frac{121}{49}$
  • Correct answer
    D
    $\frac{122}{49}$

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 $\begin{aligned} & l=\frac{11}{7}, \mathrm{~m}=\frac{-1}{7} \\ & l^2+\mathrm{m}^2\left(\frac{11}{7}\right)^2+\left(\frac{-1}{7}\right)^2=\frac{122}{49} \end{aligned}$

Hence, the correct answer is option: 'D'.

Relevant Questions