R D Sharma Solutions for Chapter: Straight Line in Space, Exercise 2: EXERCISE

Find the equations of the line passing through the point $(-1,2,1)$ and parallel to the line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}$

Find the equation of the line passing through the point $(2, -1, 3)$ and parallel to the line $\overrightarrow{r}=\left(\hat{i}-2\hat{j}+\hat{k}\right)+\lambda \left(2\hat{i}+3\hat{j}-5\hat{k}\right).$

Find the equation of the line passing through the point $\hat{i}+\hat{j}-3\hat{k}$ and perpendicular to the lines $\overrightarrow{r}=\hat{i}+\lambda (2\hat{i}+\hat{j}-3\hat{k})$ and $\overrightarrow{r}=(2\hat{i}+\hat{j}-\hat{k})+\mu (\hat{i}+\hat{j}+\hat{k})$

Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\mathrm{and} \frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

Find the vector equation of the line passing through the point $(2,-1,-1)$ which is parallel to the line $6 x-2=3 y+1=2 z-2$

If the lines $\frac{x-1}{-1}=\frac{y-2}{2\lambda }=\frac{z-3}{2}$ and $\frac{x-1}{3\lambda }=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular, find the value of $\lambda$.

Find the value of $\lambda$ so that the following lines are perpendicular to each other. $\frac{x-5}{5\lambda +2}=\frac{2-y}{5}=\frac{1-z}{-1},\frac{x}{1}=\frac{2y+1}{4\lambda }=\frac{1-z}{-3}$

Find the direction cosines of the line $\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$ Also, find the vector equation of the line through the point $A(-1, 2, 3)$ and parallel to the given line.