Question

Find the shortest distance between the lines \(\vec r=\hat i+2\hat j-4\hat k+\lambda(2\hat i+3\hat j+6\hat k)\) and \(\vec r=3\hat i+3\hat j-5\hat k+\mu(2\hat i+3\hat j+6\hat k)\).

Hint:

Same direction vector \(\Rightarrow\) parallel lines; use \(d=\dfrac{|(\vec b_2-\vec b_1)\times\vec d|}{|\vec d|}\).

Answer 1

Step 1: Understanding the Concept:
Both lines have the same direction vector \(\vec d=(2,3,6)\), so the lines are parallel. For parallel lines, the shortest distance is \(d=\dfrac{|(\vec b_2-\vec b_1)\times\vec d|}{|\vec d|}\), where \(\vec b_1,\vec b_2\) are points on each line.

Step 2: Computing the connecting vector:
\(\vec b_1=(1,2,-4)\), \(\vec b_2=(3,3,-5)\). \(\vec b_2-\vec b_1=(2,1,-1)\).

Step 3: Computing the cross product:
\((2,1,-1)\times(2,3,6)=\big(1\cdot6-(-1)\cdot3,\ -[2\cdot6-(-1)\cdot2],\ 2\cdot3-1\cdot2\big)=(9,-14,4)\).

Step 4: Computing magnitudes:
\(|(9,-14,4)|=\sqrt{81+196+16}=\sqrt{293}\). \(|\vec d|=\sqrt{4+9+36}=\sqrt{49}=7\).

Final Answer:
Shortest distance \(=\dfrac{\sqrt{293}}{7}=\boxed{\dfrac{\sqrt{293}}{7}}\) units.