Chapter 12, Problem 25CLT

(a)

To determine

To find : set of parametric equations of the line.

Answer to Problem 25CLT

  $x=-1+2t,y=2-4t,z=t$

Explanation of Solution

Given information: The line passes through the point (-1,2,0) and perpendicular to $2x-4y+z=8$ .

Calculation:

The direction ratios of the normal to the plane $2x-4y+z=8$ are $\langle 2,-4,1\rangle$ .

Since the line is perpendicular to the plane $2x-4y+z=8$ , so the direction vector of the line is $\langle 2,-4,1\rangle$ .

We know that the parametric equation of line passing through $\left(x_1,y_1,z_1\right)$ and having direction vector $\langle a,b,c\rangle$is

  $x=x_1+at,y=y_1+bt,z=z_1+ct$

Therefore, the parametric equation of line passing through (-1,2,0) and having direction vector $\langle 2,-4,1\rangle$ is

  $\begin{array}{l}x=-1+2t,y=2-4t,z=0+1t\\ x=-1+2t,y=2-4t,z=t\end{array}$

(b)

To determine

To find : set of symmetric equations of the line

Answer to Problem 25CLT

  $\frac{x+1}{2}=\frac{y-2}{-4}=\frac{z}{1}$

Explanation of Solution

Given information: The line passes through the point (-1,2,0) and perpendicular to $2x-4y+z=8$ .

Calculation:

We know that the symmetric equation of line passing through $\left(x_1,y_1,z_1\right)$ and having direction vector $\langle a,b,c\rangle$is

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

Therefore, the parametric equation of line passing through (-1,2,0) and having direction vector $\langle 2,-4,1\rangle$ is

  $\begin{array}{l}\frac{x-\left(-1\right)}{2}=\frac{y-2}{-4}=\frac{z-0}{1}\\ \frac{x+1}{2}=\frac{y-2}{-4}=\frac{z}{1}\end{array}$