Chapter 11.4, Problem 9E

(a)

To determine

The set of parametric equations

Answer to Problem 9E

The parametric equations are $x=2+2t,y=-3-3tandz=5+t$

Explanation of Solution

Given information:

Symmetric equations of the line passing through the point and parallel to the specified vector or line

  $\left(2,-3,5\right)x=5+2t,y=7-3t,z=-2+t$

Formula used:

  $\begin{array}{l}x=x_1+at\\ y=y_1+bt\\ z=z_1+ct\end{array}$

Calculation:

A vector $v=\langle a,b,c\rangle$ parallel to line L which is passing through the point $\left(x_1,y_1,z_1\right)$ then the set of parametric equations for the line is as follows,

  $\begin{array}{l}x=x_1+at\\ y=y_1+bt\\ z=z_1+ct\end{array}$

And the set of symmetric equations for the line is:

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

Consider the point (-2, 0, 3) which is parallel to the specified line $x=5+2t,y=7-3t,z=-2+t$

To find a set of parametric equations of a line, use coordinates $x_1=2,y_1=-3\text{and}{z}_1=5$

And the direction numbers are $a=2,b=-3\text{and}c=1$

Therefore,

  $\begin{array}{l}x=x_1+at\\ =2+2t\\ y=y_1+bt\\ =-3-3t\\ z=z_1+ct\\ =5+t\end{array}$

Hence, the parametric equations are $x=2+2t,y=-3-3tandz=5+t$

Conclusion:

The parametric equations are $x=2+2t,y=-3-3tandz=5+t$

(b)

To determine

The set of symmetric equations of the line passing through the point and parallel to the vector.

Answer to Problem 9E

The setof symmetric equations for the line are $\frac{x-2}{2}=\frac{y+3}{-3}=z-5$

Explanation of Solution

Given information:

Symmetric equations of the line passing through the point and parallel to the specified vector or line

  $\left(2,-3,5\right)x=5+2t,y=7-3t,z=-2+t$

Formula used:

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

Calculation:

To find the set of symmetric equations for the line,

Since a, b and c are all non-zero, therefore consider,

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

Therefore,

  $\begin{array}{l}\frac{x-2}{2}=\frac{y-\left(-3\right)}{-3}\\ =\frac{z-5}{1}\\ \frac{x-2}{2}=\frac{y+3}{-3}\\ =z-5\end{array}$

Hence, the setof symmetric equations for the line are $\frac{x-2}{2}=\frac{y+3}{-3}=z-5$

Conclusion:

The setof symmetric equations for the line are $\frac{x-2}{2}=\frac{y+3}{-3}=z-5$