The shortest distance between the lines r =((1)/(3) i +2 j +(8)/(3)...

The given lines are $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$, where:

$\vec{a}_1 = \dfrac{1}{3}\hat{i} + 2\hat{j} + \dfrac{8}{3}\hat{k}$

$\vec{b}_1 = 2\hat{i} - 5\hat{j} + 6\hat{k}$

$\vec{a}_2 = -\dfrac{2}{3}\hat{i} - \dfrac{1}{3}\hat{k}$

$\vec{b}_2 = \hat{j} - \hat{k}$

The shortest distance between two skew lines is given by $d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.

First, finding the vector $\vec{a}_2 - \vec{a}_1$:

$\vec{a}_2 - \vec{a}_1 = \left(-\dfrac{2}{3} - \dfrac{1}{3}\right)\hat{i} + (0 - 2)\hat{j} + \left(-\dfrac{1}{3} - \dfrac{8}{3}\right)\hat{k} = -\hat{i} - 2\hat{j} - 3\hat{k}$

Next, finding the cross product $\vec{b}_1 \times \vec{b}_2$:

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -5 & 6 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i}(5 - 6) - \hat{j}(-2 - 0) + \hat{k}(2 - 0) = -\hat{i} + 2\hat{j} + 2\hat{k}$

The magnitude of the cross product is:

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$

Now, finding the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (-\hat{i} + 2\hat{j} + 2\hat{k})$

$= (-1)(-1) + (-2)(2) + (-3)(2) = 1 - 4 - 6 = -9$

Finally, the shortest distance is:

$d = \dfrac{|-9|}{3} = \dfrac{9}{3} = 3$

Answer: $3$