Question

In Problems 1 to 5, all lines are in the $\left(x,y\right)$ plane.

4. Write, in parametric form, the equation of the straight line that is perpendicular to $\mathit{r}\mathbf{=}\left(2i+4j\right)\mathbf{+}\left(i-2j\right)\mathit{t}$and goes through $\left(1,0\right)\mathbf{.}$

Short answer

The parametric form of the equation is $r=i+(2i+j)t$.

Step-by-step solution

Concept of slop of the line

Slope of the line of the form $\mathit{a}\mathit{i}\mathbf{+}\mathit{b}\mathit{j}$is$\mathit{m}\mathbf{=}\frac{\mathbf{b}}{\mathbf{a}}$ .

Product of slopes of two perpendicular lines is ${\mathit{m}}_{\mathbf{1}}{\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}$.

Determine the parametric form

The equation of line is $r=(2i+4j)+(i-2j)t$ and passing through$(1,0)$.

Let $L=(i-2j)$.

The slope of line $L$ is $m=\frac{-2}{1}=-2$.

Then the slope of line perpendicular to $L$ is $\frac{-1}{m}=\frac{1}{2}$.

Then the equation of line perpendicular to $L$ is $L_1=(2i+j)$.

The line passing through $(1,0)$ and parallel to $L_1=(2i+j)$ is,

$\frac{1}{2}=\frac{y-0}{x-1}$ $\left(\because \frac{y-y_1}{x-x_1}=m\right)$

$\frac{x-1}{2}=y$

.

Hence, the parametric form of the equation is $r=i+(2i+j)t$.