012. Perpendicular Relation Between a Line and a Plane

Given a sphere SS with a radius of 22 and two lines ll and mm which are not parallel, let AA and BB be the intersection points between SS and ll, and let PP and QQ be the intersection points between SS and mm as shown below. When APQAPQ is an equilateral triangle with side 232\sqrt{3}, AB=22\overline{AB} = 2\sqrt{2}, and ABQ=π/2\angle ABQ = \pi / 2, find cosθ\cos \theta where θ\theta is the angle between the plane APBAPB and the plane APQAPQ.


With the given conditions, Pythagorean theorem, and the fact that AA, BB, PP, and QQ are on SS, the lengths of most lines are revealed. Let OO be the origin of SS. Also, let MM be the midpoint of AQ\overline{AQ}. Then, it can be found that the equilateral triangle APQAPQ includes OO since OO is the centroid of this triangle. Besides, BMOBMO is a right triangle by Pythagorean theorem.


Here is the most important observation. The line PMPM is perpendicular to the line AQAQ and BMBM. When a line ll is perpendicular some two different lines which means they are not parallel, a plane containing these two lines is also perpendicular to ll.


This fact implies that the line PMPM is perpendicular to ABQABQ. That is, APBAPB is projected into APHAPH where HH is the perpendicular from BB to the line ACAC. Therefore, cosθ\cos \theta is the ratio of these two triangles’ areas. For APBAPB, BP=23\overline{BP} = 2\sqrt{3} by Pythagorean theorem of BMPBMP, so APBAPB is an isosceles triangle such that PA=PB\overline{PA} = \overline{PB}. Area of APB=1222(23)2(2)2=25\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 APHAPH, find BH\overline{BH} first. From the area of ABQABQ, Area of ABQ=12222=1223BH    BH=263\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 APHAPH is Area of APH=123(22)2(263)2=23\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, cosθ=23/(25)=15/5\cos \theta = 2\sqrt{3} / (2\sqrt{5}) = \sqrt{15}/5.

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