Question

If the line$\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $lx+my-z=$9, then find the value of $l^2+m^2$.

Answer 1

The line $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ lies in the plane Ax + By + Cz + D = 0 iff (i) $A{x}_1+B{y}_1+C{z}_1+D=0$ and (ii) $aA+bB+cC=0$.

It is given that the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $lx+my-z=9$.

$$\begin{gathered} \therefore l\times 3+m\times \left(-2\right)-\left(-4\right)=9 \\ \Rightarrow 3l-2m=5.....\left(1\right) \end{gathered}$$
Also,

$$\begin{gathered} 2\times l+\left(-1\right)\times m+3\times \left(-1\right)=0 \\ \Rightarrow 2l-m=3.....\left(2\right) \end{gathered}$$
Solving (1) and (2), we get

l = 1 and m = −1

∴ l² + m² = 1² + (−1)² = 1 + 1 = 2

Thus, the value of l² + m² is 2.