0
$\begingroup$

Question: when $A$,$B$,$C$ and $D$ meet the following conditions, find the area of the projection of $ABCD$ onto the $yz$ plane.

  1. $A$ is $(-2,-2,-4)$.
  2. Four points $A$,$B$,$C$ and $D$ are on the sphere whose center is the origin and radius is $2\sqrt6$.
  3. $ABCD$ forms a square and the acute angle between $ABCD$ and the $xy$ plane is $cos^{-1}\frac23$.
  4. The acute angle between $\overline{AB}$ and the $x$-axis is $\frac\pi 3$.

I couldn't try much. All I was thinking was to find $C$ as an intersection among the sphere and two cones - one cone for the 3rd condition and the other for the 4th condition. But I couldn't transform the 4th condition into a condition for $\overline{AC}$.

1 Answers 1

0

You could try to solve algebraically. Suppose you want to find point $B$ and $C$, each of them with three coordinates, so you would need 6 equations. You have two equations from the fact that the points are on the same sphere: $$x_{B,C}^2+y_{B,C}^2+z_{B,C}^2=(2\sqrt{6})^2=24$$ The third and fourth equation come from the fact that $AB$ and $BC$ are sides of a square. One equation concerns the angle (must be $90^\circ$). You can write it as $\vec{AB}\cdot\vec{BC}=0$. The other equation is the lengths of $AB$ and $BC$ are the same.

The last two equations are given by the conditions 3 and 4. The vector perpendicular to the square is $\vec{AB}\times\vec{BC}$, and the angle between this vector and the $z$ axis is $\cos^{-1}\frac{2}{3}$.

Note that you might have multiple solutions.