Chapter 10.4, Problem 11E

(a)

To determine

To find: The set of parametric equations for the line through the point $\left(3,6,-5\right)$ and parallel to the lines $x=5+2t$ , $y=7-3t$ and $z=-2+t$ .

Answer to Problem 11E

The set of parametric equations for the line through the point $\left(3,6,-5\right)$ and parallel to the lines $x=5+2t$ , $y=7-3t$ and $z=-2+t$ is $x=3+2t$ , $y=6-3t$ and $z=-5+t$ .

Explanation of Solution

Given information:

The line passes through the point $\left(3,6,-5\right)$ and parallel to the lines $x=5+2t$ , $y=7-3t$ , and $z=-2+t$ .

Calculation:

The general set of equations of line in parametric form is $x=x_1+at$ , $y=y_1+bt$ and $z=z_1+ct$ , where $\left(x_1,y_1,z_1\right)$ is the passing point and $\langle a,b,c\rangle$ are the direction numbers.

Compare the line $x=3+2t$ , $y=6-3t$ and $z=-5+t$ with the general equation of line, it gives the value of $a$ , $b$ , $c$ as $2$ , $-3$ , $1$ respectively.

Both the lines are parallel. So, the direction numbers of both the lines are same. So, the direction numbers of the lines are $\langle 2,-3,1\rangle$ .

Substitute $2$ , $-3$ , $1$ for $a$ , $b$ , $c$ and $3$ , $6$ , $-5$ for $x_1$ , $y_1$ , $z_1$ in the general equation.

  $\begin{array}{l}x=x_1+at\\ x=3+2t\end{array}$ ,

  $\begin{array}{c}y=y_1+bt\\ y=6-3t\end{array}$ ,

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

Therefore, the set of parametric equations for the line through the point $\left(3,6,-5\right)$ and parallel to the lines $x=5+2t$ , $y=7-3t$ and $z=-2+t$ is $x=3+2t$ , $y=6-3t$ and $z=-5+t$ .

(b)

To determine

To find: The set of symmetric equations for the line through the point $\left(3,6,-5\right)$ and parallel to the lines $x=5+2t$ , $y=7-3t$ and $z=-2+t$ .

Answer to Problem 11E

The set of symmetric equation for the line through the point $\left(3,6,-5\right)$ and parallel to the lines $x=5+2t$ , $y=7-3t$ and $z=-2+t$ is $\frac{x-3}{2}=\frac{y-6}{-3}=\frac{z+5}{1}$ .

Explanation of Solution

Given information:

The line passes through the point $\left(3,6,-5\right)$ and parallel to the lines $x=5+2t$ , $y=7-3t$ and $z=-2+t$ .

Calculation:

The given set of parallel lines is $x=5+2t$ , $y=7-3t$ and $z=-2+t$ .

Change the set of parametric lines to set of symmetric lines.

  $\begin{array}{c}x=5+2t\\ x-5=2t\\ t=\frac{x-5}{2}\end{array}$ ,

  $\begin{array}{l}y=7-3t\\ y-7=-3t\\ t=\frac{y-7}{-3}\end{array}$ ,

  $\begin{array}{l}z=-2+t\\ z+2=t\\ t=\frac{z+2}{1}\end{array}$

So, the set of symmetric parallel lines is $\frac{x-5}{2}=\frac{y-7}{-3}=\frac{z+2}{1}$ .

Compare this set of symmetric lines to the general set of symmetric lines $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ . This gives the direction numbers of the parallel line as $\langle 2,-3,1\rangle$ . Both the lines are parallel. So, the direction numbers of both the lines are same. So, the direction numbers of the lines are $\langle 2,-3,1\rangle$ .

Substitute $2$ , $-3$ , $1$ for $a$ , $b$ , $c$ and $3$ , $6$ , $-5$ for $x_1$ , $y_1$ , $z_1$ in the general equation.

  $\begin{array}{c}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \frac{x-3}{2}=\frac{y-6}{-3}=\frac{z-\left(-5\right)}{1}\\ \frac{x-3}{2}=\frac{y-6}{-3}=\frac{z+5}{1}\end{array}$

Therefore, the set of symmetric equation for the line through the point $\left(3,6,-5\right)$ and parallel to the lines $x=5+2t$ , $y=7-3t$ and $z=-2+t$ is $\frac{x-3}{2}=\frac{y-6}{-3}=\frac{z+5}{1}$ .