Question

Find the shortest distance between the skew lines whose vector equations are given.

Hint:

Shortest distance between skew lines = Scalar triple product ÷ magnitude of cross product.

Answer 1

Concept: The shortest distance between two skew lines in 3D space is the length of the perpendicular between them. If the lines are: \[ \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \quad \text{and} \quad \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \] Then the shortest distance is given by: \[ \text{Distance} = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|} {|\vec{b}_1 \times \vec{b}_2|} \] Step 1: Identify components.
From the given vector equations: \begin{itemize} \item \( \vec{a}_1, \vec{a}_2 \) = position vectors of points on each line \item \( \vec{b}_1, \vec{b}_2 \) = direction vectors of the lines \end{itemize}
Step 2: Find vector joining the points.
\[ \vec{a}_2 - \vec{a}_1 \]
Step 3: Find cross product of direction vectors.
\[ \vec{b}_1 \times \vec{b}_2 \] This gives a vector perpendicular to both lines.
Step 4: Use scalar triple product.
\[ |(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)| \]

Step 5: Divide by magnitude of cross product.
\[ \text{Shortest distance} = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|} {|\vec{b}_1 \times \vec{b}_2|} \] Conclusion:
Use the above formula by substituting the given vectors to obtain the shortest distance.