Solid Geometry – Problem 10: Prove that , and find the dihedral angle between the two planes

Two lines $l_1$ and $l_2$ intersect at point $O$. Planes $\alpha$ and $\beta$ satisfy $$l_1\parallel \alpha,\quad l_1\parallel \beta,\quad l_2\parallel \alpha,\quad l_2\parallel \beta.$$ (Here $l_2$ is a transversal line intersecting $l_1$.)

Prove that $\alpha\parallel\beta$, and find the dihedral angle between the two planes.

Set up a 3D rectangular coordinate system with $O$ as the origin. Take $l_1$ as the $x$-axis and $l_2$ as the $y$-axis. Then direction vectors can be chosen as $$\vec u=(1,0,0)\ \text{for }l_1,\qquad \vec v=(0,1,0)\ \text{for }l_2.$$ The condition $l_1\parallel\alpha$ means $\vec u$ is parallel to plane $\alpha$, and $l_2\parallel\alpha$ means $\vec v$ is parallel to plane $\alpha$. Since $\vec u$ and $\vec v$ are not parallel, the direction space of plane $\alpha$ contains two independent directions $(1,0,0)$ and $(0,1,0)$. Therefore a normal vector of plane $\alpha$ is parallel to $$\vec n_\alpha=\vec u\times\vec v=(0,0,1).$$ Similarly, from $l_1\parallel\beta$ and $l_2\parallel\beta$, a normal vector of plane $\beta$ is parallel to $$\vec n_\beta=\vec u\times\vec v=(0,0,1).$$ Hence $\vec n_\alpha\parallel\vec n_\beta$, so $$\alpha\parallel\beta.$$ The dihedral angle between two parallel planes is $$0^\circ.$$

$\alpha\parallel\beta$, so the dihedral angle between them is $0^\circ$.