Chapter 11, Problem 9CT

(a)

To determine

To find: a set of parametric equations of the line .

Answer to Problem 9CT

  $x=8-2t,y=-2+6t,z=5-6t$

Explanation of Solution

Given information : The line passes through points A(8,-2,5) and B (6,4,-1)

Calculation:

We know that the parametric equation of line passing through $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is

  $x=x_1+t\left(x_2-x_1\right),y=y_1+t\left(y_2-y_1\right),z=z_1+t\left(z_2-z_1\right)$

Therefore, parametric equation of line passing through the points A(8,-2,5) and B(6,4,-1) is

  $\begin{array}{l}x=8+t\left(6-8\right),y=-2+t\left(4+2\right),z=5+t\left(-1-5\right)\\ x=8-2t,y=-2+6t,z=5-6t\end{array}$

(b)

To determine

To find: a set of symmetric equation of the line.

Answer to Problem 9CT

  $\frac{x-8}{1}=\frac{y+2}{-3}=\frac{z-5}{3}$

Explanation of Solution

Given information : The line passes through points A(8,-2,5) and B (6,4,-1)

Calculation:

We know that the symmetric equation of line passing through $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is

  $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Therefore, symmetric equation of line passing through the points A(8,-2,5) and B(6,4,-1) is

  $\begin{array}{l}\frac{x-8}{6-8}=\frac{y+2}{4+2}=\frac{z-5}{-1-5}\\ \frac{x-8}{-2}=\frac{y+2}{6}=\frac{z-5}{-6}\\ \frac{x-8}{1}=\frac{y+2}{-3}=\frac{z-5}{3}\end{array}$