Find the shortest distance between the skew lines (displaystyle \frac{x-2}{4}=\frac{y+1}{-3}=\frac{z-5}{2}) and (displaystyle \frac{x-0}{1}=\frac{y-4}{2}=\frac{z+1}{-1}).
Answer
(\frac{34}{sqrt{158}})
The correct answer is (\frac{34}{sqrt{158}}) — Shortest distance: (\frac{|,overrightarrow{P_1P_2}cdot(mathbf{v}_1\times\mathbf{v}_2),|}{|mathbf{v}_1\times\mathbf{v}_2|}=\frac{34}{sqrt{158}}).
Step-by-step solution
- Direction vector of (L_1): (langle4,-3,2\rangle).
- Direction vector of (L_2): (langle1,2,-1\rangle).
- Choose points (P_1(2,-1,5)) on (L_1) and (P_2(0,4,-1)) on (L_2).
- Compute cross product: (langle4,-3,2\rangle\times\langle1,2,-1\rangle=langle-1,6,11\rangle).
- Magnitude of cross product: (sqrt{(-1)^2+6^2+11^2}=sqrt{158}).
- Vector between points: (overrightarrow{P_1P_2}=(-2,5,-6)).
- Dot with cross product: ((-2)(-1)+5(6)+(-6)(11)=34) (absolute value).
- Shortest distance: (\frac{|,overrightarrow{P_1P_2}cdot(mathbf{v}_1\times\mathbf{v}_2),|}{|mathbf{v}_1\times\mathbf{v}_2|}=\frac{34}{sqrt{158}}).
Common mistake: Forgetting the absolute value or using the wrong vector difference
Concept tested: Shortest distance between skew lines
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