Question

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

Hint:

Use the shortest-distance-between-skew-lines formula with the cross product of the direction vectors.

Answer 1

Step 1: Key Formula for skew lines:
For lines \(\vec r=\vec a_1+\lambda\vec d_1\) and \(\vec r=\vec a_2+\mu\vec d_2\), the shortest distance is \(\dfrac{\big|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)\big|}{|\vec d_1\times\vec d_2|}\).

Step 2: Identifying vectors:
\(\vec a_1=(1,1,0)\), \(\vec d_1=(2,-1,1)\); \(\vec a_2=(2,1,-1)\), \(\vec d_2=(3,-5,2)\). So \(\vec a_2-\vec a_1=(1,0,-1)\).

Step 3: Computing the cross product:
\(\vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&1\\3&-5&2\end{vmatrix}=\hat i[(-1)(2)-(1)(-5)]-\hat j[(2)(2)-(1)(3)]+\hat k[(2)(-5)-(-1)(3)]=(3,-1,-7)\).

Step 4: Computing the dot product and magnitude:
\((\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=(1)(3)+(0)(-1)+(-1)(-7)=3+7=10\). \(|\vec d_1\times\vec d_2|=\sqrt{9+1+49}=\sqrt{59}\).

Final Answer:
\[ \boxed{\text{Shortest distance}=\dfrac{10}{\sqrt{59}}=\dfrac{10\sqrt{59}}{59}\text{ units}} \]