Embibe Experts Solutions for Exercise 3: EXERCISE 12.3

The shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is

The vector equation of the line of shortest distance between the lines given by $\overrightarrow{r}=\left(3\hat{i}+8\hat{j}+3\hat{k}\right)+\lambda \left(3\hat{i}-\hat{j}+\hat{k}\right)$ and $\overrightarrow{r}=\left(-3\hat{i}-7\hat{j}+6\hat{k}\right)+\mu \left(-3\hat{i}+2\hat{j}+4\hat{k}\right)$ is given by

The shortest distance between the lines $\overrightarrow{r}=\overrightarrow{a}+k\overrightarrow{b}$ and $\overrightarrow{r}=\overrightarrow{a}+l\overrightarrow{c}$ is (where $\overrightarrow{b}$ and $\overrightarrow{c}$ are non-collinear)

The lines $\overrightarrow{r}=\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)t+3\hat{\mathrm{k}}$ and $\overrightarrow{r}=\left(\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)s+3\hat{\mathrm{k}}$

The shortest distance between the lines $\overrightarrow{r}={\overrightarrow{p}}_1+t{\overrightarrow{q}}_1$ and $\overrightarrow{r}={\overrightarrow{p}}_2+s{\overrightarrow{q}}_2$ is

The point on the line $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+3}{-2}=\frac{\mathrm{z}+5}{-2}$ at a distance of $6$ from the point $\left(2,\mathrm{ }-3,-\mathrm{ }5\right)$ is

The lines $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{-1}$ and $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{4}$ are