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 $l^2 + m^2$ is equal to

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

Answer 1

Answer: (D)

$\frac{122}{49}$

Answer 2

Here's how to solve the problem:

1. Direction Vector and Point on the Line:

The given line has direction vector $\vec{d} = (2, 1, 3)$ and passes through the point $P(3, -2, -4)$.

2. Normal Vector of the Plane:

The plane $lx + my - z = 9$ has a normal vector $\vec{n} = (l, m, -1)$.

3. Conditions for the Line to Lie in the Plane:

• The direction vector of the line must be perpendicular to the normal vector of the plane: $\vec{d} \cdot \vec{n} = 0$, which gives us $2l + m - 3 = 0$, or $2l + m = 3$.
• The point $P$ on the line must satisfy the equation of the plane: $l(3) + m(-2) - (-4) = 9$, which simplifies to $3l - 2m = 5$.

4. Solving the System of Equations:

From the first equation, we have $m = 3 - 2l$. Substituting this into the second equation gives:

$3l - 2(3 - 2l) = 5$

$3l - 6 + 4l = 5$

$7l = 11$

$l = \frac{11}{7}$

Now, substitute $l$ back into the equation for $m$:

$m = 3 - 2(\frac{11}{7}) = 3 - \frac{22}{7} = \frac{21 - 22}{7} = -\frac{1}{7}$

5. Calculating $l^2 + m^2$:

$l^2 + m^2 = (\frac{11}{7})^2 + (-\frac{1}{7})^2 = \frac{121}{49} + \frac{1}{49} = \frac{122}{49}$

Therefore, $l^2 + m^2 = \frac{122}{49}$.