Question

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

Hint:

Use d = |(a2-a1).(b1 x b2)| / |b1 x b2| for two skew lines in vector form.

Answer 1

Step 1: Identify \(\vec a_1,\vec b_1,\vec a_2,\vec b_2\):
\(\vec a_1=(1,2,1)\), \(\vec b_1=(1,-1,1)\), \(\vec a_2=(2,-1,-1)\), \(\vec b_2=(2,1,2)\).

Step 2: Compute \(\vec a_2-\vec a_1\):
\[ \vec a_2-\vec a_1=(1,-3,-2) \]

Step 3: Compute \(\vec b_1\times\vec b_2\):
\[ \vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix}=\hat i(-2-1)-\hat j(2-2)+\hat k(1+2)=(-3,0,3) \]

Step 4: Compute the shortest distance formula:
\[ d=\dfrac{\big|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)\big|}{|\vec b_1\times\vec b_2|} \]
Numerator: \((1)(-3)+(-3)(0)+(-2)(3)=-3+0-6=-9\), so \(|-9|=9\).
Denominator: \(|(-3,0,3)|=\sqrt{9+0+9}=\sqrt{18}=3\sqrt2\).

Final Answer:
\[ d=\dfrac9{3\sqrt2}=\dfrac3{\sqrt2}=\dfrac{3\sqrt2}2 \] \[ \boxed{d=\dfrac{3\sqrt2}2} \]