The Shortest Distance Between Skew Lines and its Equation

Let $Q$ be the cube with the set of vertices $\left\{\left(x_1, x_2, x_3\right)\in {\mathrm{ℝ}}^3 : x_1, x_2, x_3\in \left\{0, 1\right\}\right\}$. Let $F$ be the set of all twelve lines containing the diagonals of the six faces of the cube $Q$. Let $S$ be the set of all four lines containing the main diagonals of the cube $Q$; for instance, the line passing through the vertices $\left(0, 0, 0\right)$ and $\left(1, 1, 1\right)$ is in $S$. For lines ${\ell }_1$ and ${\ell }_2$, let $d\left({\ell }_1, {\ell }_2\right)$ denote the shortest distance between them. Then the maximum value of $d\left({\ell }_1, {\ell }_2\right)$, as ${\ell }_1$ varies over $F$ and ${\ell }_2$ varies over $S$, is

Find the shortest distance between the lines $\overrightarrow{r}=6\hat{i}-\hat{j}-4\hat{k}+\lambda \left(2\hat{i}-\hat{j}-\hat{k}\right)$ and $\overrightarrow{r}=3\hat{i}+2\hat{j}+5\hat{k}+\mu \left(\hat{i}+2\hat{j}+3\hat{k}\right)$.

Find the shortest distance between the lines $\overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})$and $\overrightarrow{r}=5\hat{i}-2\hat{j}+\mu (3\hat{i}+2\hat{j}+6\hat{k})$.

If the lines intersect find their point of intersection.

consider the lines $L_1:\frac{x-1}{1}=\frac{y}{-1}=\frac{z}{1}; L_2:\frac{x}{1}=\frac{y+2}{-2}=\frac{z-1}{1}$ and the plane $\pi :x-y-2z-1=0$

The value of $\alpha$ for which the shortest distance between the lines represented by $y+z=0, z+x=0$ and $x+y=0, x+y+z=\alpha$ is $1,$ is

If the shortest distance between the lines $x+2\lambda =2y=-12z, x=y+4\lambda =6z-12\lambda$ is $4\sqrt{2}$ units, then value of $\lambda$ is

Find the shortest distance between the pair of lines:

$\frac{x-3}{2}=\frac{y-4}{1}=\frac{z+1}{-3}$ and $\frac{{\displaystyle x-1}}{{\displaystyle -1}}=\frac{{\displaystyle y-3}}{{\displaystyle 3}}=\frac{{\displaystyle z-1}}{{\displaystyle 2}}$

$OABC$ is a tetrahedron in with $O$ as the origin and position vectors of points $A, B, C$ as $\hat{i}+2\hat{j}+3\hat{k}, 2\hat{i}+\alpha \hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}+2\hat{k}$ respectively, then the integral value of $\alpha$ to have shortest distance between $\overrightarrow{\mathrm{OA}}$ $\&$ $\overrightarrow{\mathrm{BC}}$ as $\sqrt{\frac{3}{2}}$, is

Consider the two given lines, $L_1:\frac{x+2}{1}=\frac{y+1}{3}=\frac{z+1}{2}$ and $L_2:\frac{x+2}{2}=\frac{y-2}{1}=\frac{z-3}{3}$ . Let $\frac{a}{b\sqrt{3}}$ be the shortest distance between lines $L_1$ and $L_2,$ then find the value of $(a-2b)$ ? (where $a$ and $b$ are co-prime numbers)

Consider two lines in space as $L_1 : \overrightarrow{r_1}=\hat{\mathrm{j}}+2\hat{\mathrm{k}}+\lambda \left(3\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)$ and $L_2 : \overrightarrow{r_2}=4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+6\hat{\mathrm{k}}+\mu \left(\hat{\mathrm{i}}+2\hat{\mathrm{k}}\right)$ where $\lambda , \mu \in R.$ If the shortest distance between these lines is $\sqrt{d}.$ Then the value of $d$ will be:

Find the shortest distance between the lines given by

$\overline{r}=(8+3\lambda )\hat{i}-(9+16\lambda )\hat{j}+(10+7\lambda )\hat{k}\text{ and }\overline{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})$.

If two lines $L_1=\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-2}{1}$ and $L_2=\frac{x+2}{2}=\frac{y+1}{2}=\frac{z+2}{3}$  are given, then the possible minimum distance between $L_1$ and $L_2$ is

The shortest distance between the lines $2\mathrm{x}+\mathrm{y}+\mathrm{z}-1=0=3\mathrm{x}+\mathrm{y}+2\mathrm{z}-2$ and $\mathrm{x}=\mathrm{y}=\mathrm{z}$ is:

The shortest distance between the skew lines $\frac{x+3}{-4}=\frac{y-6}{3}=\frac{z}{2}$ and $\frac{x+2}{-4}=\frac{y}{1}=\frac{z-7}{1}$ is