Question

Find the equation of line in vector and in cartesian form that passes through the point with position vector $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$

Answer 1

Step 1: Solve for the equation of the line in vector form

It is given that the line passes through the point with position vector $\overrightarrow{a}=2\hat{i}-\hat{j}+4\hat{k}$ and in the direction $\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}$

Let $\overrightarrow{r}$ is any line vector
$\therefore$Equation of a line through a point with position vector $\overrightarrow{a}$and parallel to $\overrightarrow{b}$is

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ where $\lambda$ is an arbitrary constant

So the required equation of the line in vector form is

$\Rightarrow \overrightarrow{r}=2\hat{i}-\hat{j}+4\hat{k}+\lambda \left(\hat{i}+2\hat{j}-\hat{k}\right)$

$\Rightarrow (\lambda +2)\hat{i}+(2\lambda -1)\hat{j}+(-\lambda +4)\hat{k}$

$\therefore$Equation of line in vector form is $\left(2+\lambda \right)\hat{i}+\left(2\lambda -1\right)\hat{j}+\left(4-\lambda \right)\hat{k}$

Step 2: Solve for the equation of the line in cartesian form

General equation of line in cartesian form passing through point $\left(x_{1,}{y}_1,z_1\right)$ and having direction ratio $a,b,c$ is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Since the line passes through a point with position vector $2\hat{i}-\hat{j}+4\hat{k}$

$\therefore x_1=2,y_1=-1,z_1=4$

Also line is in the direction of $\hat{i}+2\hat{j}-\hat{k}$

$\therefore$Direction ratios are $a=1,b=2,c=-1$

$$\begin{gathered} \Rightarrow \frac{x-2}{1}=\frac{y-(-1)}{2}=\frac{z-4}{-1} \\ \Rightarrow \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1} \end{gathered}$$

Hence the equations of line in vector and cartesian form are $(2+\lambda )\hat{i}+(2\lambda -1)\hat{j}+(4-\lambda )\hat{k}$ and $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}$ respectively.