Q.Find the shortest distance between the lines $\vec{r} = (\hat{i}+2\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r} = (2\hat{i}-\hat{j}-\hat{k}) + \mu(2\hat{i}+\hat{j}+2\hat{k})$. OR Find the equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to each of the planes $x+2y+3z=5$ and $3x+3y+z=0$.

Using the skew-line shortest-distance formula $\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|$ gives $\dfrac{3}{\sqrt2}$. (OR: the required plane's normal is $\vec n_1\times\vec n_2$ of the two given planes, giving $7x-8y+3z+25=0$.)

Shortest distance between the two lines

Line 1: $\vec r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)$, so $\vec a_1=(1,2,1)$, $\vec b_1=(1,-1,1)$.

Line 2: $\vec r=(2\hat i-\hat j-\hat k)+\mu(2\hat i+\hat j+2\hat k)$, so $\vec a_2=(2,-1,-1)$, $\vec b_2=(2,1,2)$.

For skew lines, shortest distance $=\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|$.

$\vec a_2-\vec a_1 = (1,-3,-2).$

$\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix} = \hat i[(-1)(2)-(1)(1)] - \hat j[(1)(2)-(1)(2)] + \hat k[(1)(1)-(-1)(2)]$

$= \hat i(-3) - \hat j(0) + \hat k(3) = (-3,0,3).$

$|\vec b_1\times\vec b_2| = \sqrt{9+0+9}=\sqrt{18}=3\sqrt2.$

$(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2) = (1)(-3)+(-3)(0)+(-2)(3) = -3+0-6=-9.$

$\text{Shortest distance} = \left|\frac{-9}{3\sqrt2}\right| = \frac{9}{3\sqrt2} = \frac{3}{\sqrt2} = \frac{3\sqrt2}{2}.$

OR: Plane through $(-1,3,2)$ perpendicular to $x+2y+3z=5$ and $3x+3y+z=0$

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