Chapter 10, Problem 57RE

(a)

To determine

To find: The set of parametric equation.

Answer to Problem 57RE

The set of parametric equation are $x=-1+2t$ , $y=3-t$ and $z=5+3t$ .

Explanation of Solution

Given information:

The given points is $p\left(-1,3,5\right)$ and parallel to $x=1+2t$ , $y=2-t$ and $z=1+3t$ .

Calculation:

Consider the set of parametric equation of a line.

  $a=2$ , $b=-1$ and $c=3$ with the initial points $p\left(-1,3,5\right)$ .

Calculate the equation.

  $\begin{array}{c}x=x_1+at\\ =-1+2t\\ =-1+2t\end{array}$

  $\begin{array}{c}y=y_1+bt\\ =3-t\end{array}$

  $\begin{array}{c}z=z_1+ct\\ =5+3t\end{array}$

Therefore, the set of parametric equation are $x=-1+2t$ , $y=3-t$ and $z=5+3t$ .

(b)

To determine

To find: The set of symmetric equation for the line.

Answer to Problem 57RE

The symmetric sets are $\frac{x-\left(-1\right)}{2}=\frac{y-3}{-1}=\frac{z-5}{3}$ .

Explanation of Solution

Given information:

The given points is $p\left(-1,3,5\right)$ and parallel to $x=1+2t$ , $y=2-t$ and $z=1+3t$ .

Calculation:

Consider the symmetric equation of a line, since $a$ , $b$ and $c$ are all non-zero,

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

Substitute the values.

  $\frac{x-\left(-1\right)}{2}=\frac{y-3}{-1}=\frac{z-5}{3}$

Therefore, the symmetric sets are $\frac{x-\left(-1\right)}{2}=\frac{y-3}{-1}=\frac{z-5}{3}$ .