Question

A line is drawn in the direction of $(\hat{i}+\hat{j}-2\hat{k})$ and it passes through a point with the position vector $(2\hat{i}-\hat{j}-4\hat{k})$. Find the equations of the line in the vector as well as Cartesian forms.

Answer 1

$Equation\text{of a line passing through a point with position vector}\overrightarrow{\text{a}}\text{and parallel to vector}\overrightarrow{\text{b}}\text{is}$
$\overrightarrow{\text{r}}\text{=}\overrightarrow{\text{a}}\text{+}\lambda \overrightarrow{b}$
$Here$
$\overrightarrow{\text{a}}=2i-j-4k$
$\overrightarrow{b}=i+j-2k$
$Hence$
$\overrightarrow{\text{r}}\text{=}(2i-j-4k)\text{+}\lambda (i+j-2k)\left[Invectorform\right]$
$InCartesianform$
$Equation\text{of a line passing through}(x_1,y_1,z_1)\text{and parallel to a line having direction rarios a,b,c is}$
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
$Here$
$x_1=2$
$y_1=-1$
$z_1=-4$
$directionratios:a=1,b=1,c=-2$
$Hencetheequationis$
$\frac{x-\left(2\right)}{1}=\frac{y-(-1)}{1}=\frac{z-(-4)}{-2}$
$\Rightarrow \frac{x-2}{1}=\frac{y+1}{1}=\frac{z+4}{-2}$