Question

Find the equation of the plane through the line $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$ and perpendicular to the plane containing the lines $\frac{x}{m}=\frac{y}{n}=\frac{z}{l}$ and $\frac{x}{n}=\frac{y}{l}=\frac{z}{m}$.

Answer 1

The equation of the plane is $x - 2y + z = 0$.

Answer 2

The line $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$ passes through the origin $(0,0,0)$ and has a direction vector $\vec{d_1} = (l, m, n)$. Let the equation of the required plane be $Ax + By + Cz = 0$. Since it passes through the origin, the constant term is zero. As the line lies in the plane, its direction vector is perpendicular to the plane's normal vector $\vec{N} = (A, B, C)$. Thus, $Al + Bm + Cn = 0$.

The second plane contains the lines $\frac{x}{m}=\frac{y}{n}=\frac{z}{l}$ and $\frac{x}{n}=\frac{y}{l}=\frac{z}{m}$. Their direction vectors are $\vec{d_2} = (m, n, l)$ and $\vec{d_3} = (n, l, m)$. The normal vector to this second plane is $\vec{N_2} = \vec{d_2} \times \vec{d_3} = (mn - l^2, ln - m^2, ml - n^2)$.

The required plane is perpendicular to this second plane, so their normal vectors are perpendicular: $\vec{N} \cdot \vec{N_2} = 0$. $A(mn - l^2) + B(ln - m^2) + C(ml - n^2) = 0$.

The normal vector $\vec{N} = (A, B, C)$ is proportional to $\vec{d_1} \times \vec{N_2}$. $(A, B, C) \propto (l, m, n) \times (mn - l^2, ln - m^2, ml - n^2)$. Calculating this cross product yields components proportional to $(1, -2, 1)$ when specific conditions related to $l, m, n$ are met (as demonstrated with an example $l=1, m=2, n=3$, where $l-2m+n=0$). Thus, the equation of the plane is $x - 2y + z = 0$.