Chapter 10.4, Problem 12E

(a)

To determine

To find:The set of parametric equations.

Answer to Problem 12E

Theparametric equations are $x=6-3t$ , $y=8-10t$ and $z=12t$ .

Explanation of Solution

Given information:The point are $\left(6,8,0\right)$ and $\left(3,-2,12\right)$ .

Formula used: The parametric equations for the line parallel to vector $v=\langle a,b,c\rangle$ and passing through the point $\left(x_1,y_1,z_1\right)$ are given by,

  $x=x_1+at$ , $y=y_1+bt$ and $z=z_1+ct$

Calculation:

The direction vector can be obtained by the given two points as:

  $\begin{array}{c}v=\langle 3-6,-2-8,12-0\rangle \\ =\langle -3,-10,12\rangle \end{array}$

So, the normal vector is $v=\langle -3,-10,12\rangle$ and point on the plane is $\left(6,8,0\right)$ .

Substitute 6 for $x_1$ and $-3$ for $a$ in the equation $x=x_1+at$ .

  $\begin{array}{c}x=6+\left(-3\right)t\\ =6-3t\end{array}$

Substitute 8 for $y_1$ and $-10$ for $b$ in the equation $y=y_1+bt$ .

  $\begin{array}{c}y=8+\left(-10\right)t\\ =8-10t\end{array}$

Substitute 0 for $z_1$ and 12 for $c$ in the equation $z=z_1+ct$ .

  $\begin{array}{c}z=0+12t\\ =12t\end{array}$

Therefore, the parametric equations are $x=6-3t$ , $y=8-10t$ and $z=12t$ .

(b)

To determine

To find:The set of symmetric equations of the line passes through the two points $\left(6,8,0\right)$ and $\left(3,-2,12\right)$ .

Answer to Problem 12E

Theset of symmetric equation is $\frac{x-6}{-3}=\frac{y-8}{-10}=\frac{z}{12}$ .

Explanation of Solution

Given information: The points are $\left(6,8,0\right)$ and $\left(3,-2,12\right)$ .

Calculation:

The normal vector is $v=\langle -3,-10,12\rangle$ . It can be observed that the direction numbers are all non-zero.

The symmetric equations can be find as follow:

  $\begin{array}{c}\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}\\ \frac{x-6}{-3}=\frac{y-8}{-10}=\frac{z-0}{12}\\ \frac{x-6}{-3}=\frac{y-8}{-10}=\frac{z}{12}\end{array}$

Therefore, the set of symmetric equation is $\frac{x-6}{-3}=\frac{y-8}{-10}=\frac{z}{12}$ .