Question

OABCD is a pyramid with square base ABCD of unit side and vertex O such that OA = OB = OC = OD = 1, then

• A. angle between the line AC and OD is $\frac{\pi}{3}$
• B. angle between the line AC and OD is $\frac{\pi}{2}$
• C. shortest distance between AC and OD is $\frac{1}{\sqrt{2}}$
• D. shortest distance between AC and OD is $\frac{1}{2}$

Answer 1

Answer: (B), (D)

angle between the line AC and OD is $\frac{\pi}{2}$; shortest distance between AC and OD is $\frac{1}{2}$

Answer 2

Solution Outline

1. Coordinates Assignment
   Let

   $$A=(0,0,0),\quad B=(1,0,0),\quad C=(1,1,0),\quad D=(0,1,0).$$

   The center of square $ABCD$ is $M=(\tfrac12,\tfrac12,0)$.
   Since $OA=OB=OC=OD=1$, if $O=(\tfrac12,\tfrac12,h)$, then

   $$OA^2 = \bigl(\tfrac12\bigr)^2 + \bigl(\tfrac12\bigr)^2 + h^2 = 1 \;\Longrightarrow\; h^2 = \tfrac12,\;h=\tfrac{1}{\sqrt2}.$$

2. Angle Between $AC$ and $OD$
   Direction vectors:

   $$\vec{v}=\overrightarrow{AC}=(1,1,0),\quad \vec{w}=\overrightarrow{OD}=(0,1,0)-\bigl(\tfrac12,\tfrac12,\tfrac1{\sqrt2}\bigr) =\bigl(-\tfrac12,\tfrac12,-\tfrac1{\sqrt2}\bigr).$$

   Dot product:

   $$\vec{v}\cdot\vec{w}=1\cdot(-\tfrac12)+1\cdot(\tfrac12)+0\cdot(-\tfrac1{\sqrt2})=0.$$

   Hence $\cos\theta=0$, so $\theta=\tfrac{\pi}{2}$.

3. Shortest Distance Between Skew Lines
   Formula:

   $$d = \frac{\bigl|\bigl(\overrightarrow{O A}\bigr)\cdot(\vec{v}\times\vec{w})\bigr|} {\|\vec{v}\times\vec{w}\|}.$$

   Compute:

   $$\overrightarrow{OA}=(\tfrac12,\tfrac12,\tfrac1{\sqrt2}), \quad \vec{v}\times\vec{w} =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&1&0\\ -\tfrac12&\tfrac12&-\tfrac1{\sqrt2} \end{vmatrix} =\bigl(-\tfrac1{\sqrt2},\;\tfrac1{\sqrt2},\;1\bigr),$$ $$\|\vec{v}\times\vec{w}\|=\sqrt{\tfrac12+\tfrac12+1}=\sqrt2,\quad (\overrightarrow{OA})\cdot(\vec{v}\times\vec{w})=\frac1{\sqrt2}.$$

   Thus

   $$d=\frac{\tfrac1{\sqrt2}}{\sqrt2}=\frac12.$$