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

Shortest distance formula for skew lines.

$A_1(4,-1,0),\ \bar b_1=(1,2,-3)$; $A_2(1,-1,-2),\ \bar b_2=(1,4,-5)$

$A_2-A_1=(-3,0,-2)$

$\bar b_1\times\bar b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&-3\\1&4&-5\end{vmatrix}=(-10+12,-(-5+3),4-2)=(2,2,2)$

$|\bar b_1\times\bar b_2|=\sqrt{4+4+4}=2\sqrt3$

$(A_2-A_1)\cdot(\bar b_1\times\bar b_2)=-6+0-4=-10$

Distance $=\dfrac{|-10|}{2\sqrt3}=\dfrac5{\sqrt3}=\dfrac{5\sqrt3}3$

Shortest distance $=\left|\dfrac{(\bar a_2-\bar a_1)\cdot(\bar b_1\times\bar b_2)}{|\bar b_1\times\bar b_2|}\right|$.

Line 1: point $A_1(1,2,3)$, direction $\bar b_1=(2,3,4)$.

Line 2: point $A_2(2,4,5)$, direction $\bar b_2=(3,4,5)$.

$A_2-A_1=(1,2,2)$

$\bar b_1\times\bar b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&4\\3&4&5\end{vmatrix}=(15-16,-(10-12),8-9)=(-1,2,-1)$

$|\bar b_1\times\bar b_2|=\sqrt{1+4+1}=\sqrt6$

$(A_2-A_1)\cdot(\bar b_1\times\bar b_2)=(1)(-1)+(2)(2)+(2)(-1)=1$

Shortest distance $=\dfrac{|1|}{\sqrt6}=\dfrac{1}{\sqrt6}$

Use the shortest-distance-between-skew-lines formula with the connecting vector and $\vec d_1\times\vec d_2$.

Line 1: point $A_1(1,2,3)$, direction $\vec d_1=(2,3,4)$.

Line 2: point $A_2(2,4,5)$, direction $\vec d_2=(3,4,5)$.

$\vec d_1 \times \vec d_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat i(15-16) - \hat j(10-12) + \hat k(8-9) = (-1,\,2,\,-1)$

$|\vec d_1\times\vec d_2| = \sqrt{(-1)^2+2^2+(-1)^2} = \sqrt{6}$

$\vec{A_1A_2} = (2-1,\,4-2,\,5-3) = (1,2,2)$

$\vec{A_1A_2}\cdot(\vec d_1\times\vec d_2) = 1(-1)+2(2)+2(-1) = -1+4-2 = 1$

Shortest distance $= \dfrac{|\vec{A_1A_2}\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|} = \dfrac{1}{\sqrt6}$

Use the skew-lines shortest-distance formula $d=\left|\dfrac{(\bar a_2-\bar a_1)\cdot(\bar b_1\times\bar b_2)}{|\bar b_1\times\bar b_2|}\right|$.

Lines: $\bar r=(4\hat i-\hat j)+\lambda(\hat i+2\hat j-3\hat k)$ and $\bar r=(\hat i-\hat j+2\hat k)+\mu(\hat i+4\hat j-5\hat k)$.

So $\bar a_1=4\hat i-\hat j$, $\bar b_1=\hat i+2\hat j-3\hat k$, $\bar a_2=\hat i-\hat j+2\hat k$, $\bar b_2=\hat i+4\hat j-5\hat k$.

Step 1: $\bar a_2-\bar a_1=(1-4)\hat i+(-1-(-1))\hat j+(2-0)\hat k=-3\hat i+0\hat j+2\hat k$

Step 2: $\bar b_1\times\bar b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&-3\\1&4&-5\end{vmatrix}$

$=\hat i(2(-5)-(-3)(4))-\hat j(1(-5)-(-3)(1))+\hat k(1(4)-2(1))$

$=\hat i(-10+12)-\hat j(-5+3)+\hat k(4-2)=2\hat i+2\hat j+2\hat k$

$|\bar b_1\times\bar b_2|=\sqrt{2^2+2^2+2^2}=\sqrt{12}=2\sqrt3$

Step 3: $(\bar a_2-\bar a_1)\cdot(\bar b_1\times\bar b_2)=(-3)(2)+(0)(2)+(2)(2)=-6+0+4=-2$

Step 4: Shortest distance:

$d=\left|\frac{-2}{2\sqrt3}\right|=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$