Find the shortest distance between two skew lines given position vectors and magnitude conditions for a tetrahedron $ABCD$.
Step-by-Step Solution
Key Concept: The shortest distance between two skew lines is found using the cross product formula involving the direction vectors and a connecting vector.
Given $|\vec{b} - \vec{a}| = 12$ and $|\vec{d} - \vec{c}| = 15$, the shortest distance between skew lines is calculated using the formula: $\text{Shortest distance} = \frac{|(\vec{c} - \vec{a}) \cdot ((\vec{b} - \vec{a}) \times (\vec{d} - \vec{c}))|}{|(\vec{b} - \vec{a}) \times (\vec{d} - \vec{c})|} = 10$. Using the volume formula (equation 2), we find $(\vec{b} - \vec{a}) \times (\vec{d} - \vec{c}) = 6 \times 200 = 1200$. From the magnitude and angle relationship: $|\vec{b} - \vec{a}| |\vec{d} - \vec{c}| \sin \theta = 120$, giving $\sin \theta = \frac{2}{3}$ and $\theta = \arcsin(\frac{2}{3}) \approx 41.8°$.
Correct Answer: 10