Question

 If the line \[\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z + 4}{3}\]  lies in the plane  \[lx + my - z =\]   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 

\[lx + my - z = 9\]

\[\therefore l \times 3 + m \times \left( - 2 \right) - \left( - 4 \right) = 9\]
\[ \Rightarrow 3l - 2m = 5 . . . . . \left( 1 \right)\]

Also, 

\[2 \times l + \left( - 1 \right) \times m + 3 \times \left( - 1 \right) = 0\]
\[ \Rightarrow 2l - m = 3 . . . . . \left( 2 \right)\]

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.