# 012. Perpendicular Relation Between a Line and a Plane

#### Given a sphere $S$ with a radius of $2$ and two lines $l$ and $m$ which are not parallel, let $A$ and $B$ be the intersection points between $S$ and $l$, and let $P$ and $Q$ be the intersection points between $S$ and $m$ as shown below. When $APQ$ is an equilateral triangle with side $2\sqrt{3}$, $\overline{AB} = 2\sqrt{2}$, and $\angle ABQ = \pi / 2$, find $\cos \theta$ where $\theta$ is the angle between the plane $APB$ and the plane $APQ$.

With the given conditions, Pythagorean theorem, and the fact that $A$, $B$, $P$, and $Q$ are on $S$, the lengths of most lines are revealed. Let $O$ be the origin of $S$. Also, let $M$ be the midpoint of $\overline{AQ}$. Then, it can be found that the equilateral triangle $APQ$ includes $O$ since $O$ is the centroid of this triangle. Besides, $BMO$ is a right triangle by Pythagorean theorem.

Here is the most important observation. The line $PM$ is perpendicular to the line $AQ$ and $BM$. When a line $l$ is perpendicular some two different lines which means they are not parallel, a plane containing these two lines is also perpendicular to $l$.

This fact implies that the line $PM$ is perpendicular to $ABQ$. That is, $APB$ is projected into $APH$ where $H$ is the perpendicular from $B$ to the line $AC$. Therefore, $\cos \theta$ is the ratio of these two triangles’ areas. For $APB$, $\overline{BP} = 2\sqrt{3}$ by Pythagorean theorem of $BMP$, so $APB$ is an isosceles triangle such that $\overline{PA} = \overline{PB}$. \begin{aligned} \text{Area of } APB = \frac{1}{2} \cdot 2\sqrt{2} \cdot \sqrt{(2\sqrt{3})^2 - (\sqrt{2})^2} = 2\sqrt{5} \end{aligned}

For $APH$, find $\overline{BH}$ first. From the area of $ABQ$, \begin{aligned} \text{Area of } ABQ = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \frac{1}{2} \cdot 2\sqrt{3} \cdot \overline{BH} \implies \overline{BH} = \frac{2\sqrt{6}}{3} \end{aligned}

As such, the area of $APH$ is \begin{aligned} \text{Area of } APH = \frac{1}{2} \cdot 3 \cdot \sqrt{(2\sqrt{2})^2 - \left(\frac{2\sqrt{6}}{3}\right)^2} = 2\sqrt{3} \end{aligned}

Therefore, $\cos \theta = 2\sqrt{3} / (2\sqrt{5}) = \sqrt{15}/5$.