Question

Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are  $$\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$$  Also, reduce the equation obtained in vector form.

Answer 1

We know that the cartesian equation of a line passing through a point with position vector ${\displaystyle \overrightarrow{a}}$  and parallel to the vector ${\displaystyle \overrightarrow{m}}$ is  $$\frac{x-x_1}{a}=\frac{y-y_2}{b}=\frac{z-z_3}{c}$$ 

Here, 

$$\overrightarrow{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$$

$$\overrightarrow{m}=a\hat{i}+b\hat{j}+c\hat{k}$$

Here,  $$\overrightarrow{a}=\hat{i}-\hat{j}+2\hat{k}\text{ and }\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$$ 

Cartesian equation of the required line is 

$$\frac{x-1}{1}=\frac{y-(-1)}{2}=\frac{z-2}{-2}$$

$$\Rightarrow \frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}$$

We know that the cartesian equation of a line passing through a point with position vector ${\displaystyle \overrightarrow{a}}$ and parallel to the vector ${\displaystyle \overrightarrow{m}}$ is $$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{m}$$ 

Here, the line is passing through the point  (1,1,-2)  and its direction ratios are proportional to 1, 2, -2 .

Vector equation of the required line is

$$\overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda (\hat{i}+2\hat{j}-2\hat{k})$$