Q.Find the shortest distance between the lines $\vec{r} = 6\hat{i}+2\hat{j}+2\hat{k}+\lambda(\hat{i}-2\hat{j}+2\hat{k})$ and $\vec{r} = -4\hat{i}-\hat{k}+\mu(3\hat{i}-2\hat{j}-2\hat{k})$. OR Find the equations of two lines through the origin which intersect the line $\dfrac{x-3}{2}=\dfrac{y-3}{1}=\dfrac{z}{1}$ at $\dfrac{\pi}{3}$.

Main: apply the skew-line shortest-distance formula $\left|\dfrac{(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)}{|\vec b_1\times\vec b_2|}\right|$. OR: parametrize a general point on the given line, form the direction to the origin, and solve for the parameter using the given $\pi/3$ angle condition.

Main question.

$\vec r_1 = 6\hat i+2\hat j+2\hat k+\lambda(\hat i-2\hat j+2\hat k),\qquad \vec r_2 = -4\hat i-\hat k+\mu(3\hat i-2\hat j-2\hat k).$

Here $\vec a_1=(6,2,2)$, $\vec b_1=(1,-2,2)$, $\vec a_2=(-4,0,-1)$, $\vec b_2=(3,-2,-2)$.

$\vec a_2-\vec a_1 = (-10,-2,-3)$.

$\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&2\\3&-2&-2\end{vmatrix} = \hat i[(-2)(-2)-2(-2)] - \hat j[1(-2)-2(3)] + \hat k[1(-2)-(-2)(3)] = 8\hat i+8\hat j+4\hat k.$

$|\vec b_1\times\vec b_2| = \sqrt{8^2+8^2+4^2} = \sqrt{144} = 12.$

$(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1) = 8(-10)+8(-2)+4(-3) = -80-16-12 = -108.$

$\text{Shortest distance} = \frac{|-108|}{12} = 9.$

OR question. Line $L$: $\dfrac{x-3}{2}=\dfrac{y-3}{1}=\dfrac{z}{1}=t$, so a general point on $L$ is $P(t)=(3+2t,\ 3+t,\ t)$, with direction $\vec d=(2,1,1)$, $|\vec d|=\sqrt6$.

We want lines through the origin $O$ meeting $L$ at $P(t)$ such that the angle between $\vec{OP}$ and $\vec d$ is $\pi/3$, i.e.

$\cos\frac\pi3 = \frac{\vec{OP}\cdot\vec d}{|\vec{OP}||\vec d|} = \frac12.$

$\vec{OP} = (3+2t,\ 3+t,\ t)$.

$\vec{OP}\cdot\vec d = 2(3+2t)+1(3+t)+1(t) = 9+6t.$

$|\vec{OP}|^2 = (3+2t)^2+(3+t)^2+t^2 = 18+18t+6t^2.$

From $\dfrac{9+6t}{\sqrt6\,|\vec{OP}|}=\dfrac12$: $\ 2(9+6t) = \sqrt6\,|\vec{OP}|$. Squaring:

$4(9+6t)^2 = 6|\vec{OP}|^2 \implies 4(324+... )$

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