Vinod Singh and Shweta Pawar Solutions for Chapter: Line, Exercise 1: Classical Thinking

The distance of the point $2\hat{i}+\hat{j}+\hat{k}$ from the line $\overrightarrow{r}=-\hat{i}+2\hat{j}+2\hat{k}+\lambda \left(3\hat{i}+\hat{k}\right)$ is

The perpendicular distance of the point $\left(2, 4, -1\right)$ from the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ is

The length of perpendicular from $\left(1,6,3\right)$ to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is

Skew lines are

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

The shortest distance between lines $\overrightarrow{r_1}=\left(\hat{i}-\hat{j}\right)+\lambda \left(2\hat{i}+\hat{k}\right)$ and $\overrightarrow{r_2}=\left(2\hat{i}-\hat{j}\right)+\mu \left(\hat{i}-\hat{j}-\hat{k}\right)$ is

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

If the two lines $\frac{x-1}{3}=\frac{y-k}{6}=\frac{z+1}{-2}$ and $\frac{x-2}{-1}=\frac{y-2}{4}=\frac{z+1}{-1}$ intersect at a point then $k$ is