Q.The shortest distance between the lines $\vec{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+\mu(2\hat{i}+\hat{j}+2\hat{k})$ is
(A) $\dfrac{1}{\sqrt{3}}$
(B) $\dfrac{1}{\sqrt{2}}$
(C) $\dfrac{3}{\sqrt{2}}$
(D) $\dfrac{\sqrt{3}}{2}$

Shortest distance formula for skew lines.

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

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

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

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

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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 $\bar b_2=(3,4,5)$.

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

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

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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 $\vec d_2=(3,4,5)$.

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

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

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

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

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

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

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