Chapter 10.4, Problem 6E

(a)

To determine

To find:The set of parametric equations.

Answer to Problem 6E

Theparametric equations are $x=3+3t$ , $y=-5-7t$ and $z=1-10t$ .

Explanation of Solution

Given information:

The point is $\left(3,-5,1\right)$ and vector is $v=\langle 3,-7,-10\rangle$ .

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 is $v=\langle 3,-7,-10\rangle$ . So, the direction numbers are $a=3$ , $b=-7$ and $c=-10$ . The point is $\left(3,-5,1\right)$ .

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

  $x=3+3t$

Substitute $-5$ for $y_1$ and $-7$ for $b$ in the equation $y=y_1+bt$ .

  $y=-5-7t$

Substitute 1 for $z_1$ and $-10$ for $c$ in the equation $z=z_1+ct$ .

  $z=1-10t$

Therefore, the parametric equations are $x=3+3t$ , $y=-5-7t$ and $z=1-10t$ .

(b)

To determine

To find:The set of symmetric equations for the line through the point $\left(3,-5,1\right)$ and parallel to the specified vector $v=\langle 3,-7,-10\rangle$ .

Answer to Problem 6E

Theset of symmetric equation is $\frac{x-3}{3}=\frac{y+5}{-7}=\frac{z-1}{-10}$ .

Explanation of Solution

Given information:

The point is $\left(3,-5,1\right)$ and vector is $v=\langle 3,-7,-10\rangle$ .

Calculation:

Since the direction numbers are all non-zero then symmetric equation will be as follows:

  $\begin{array}{c}\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}\\ \frac{x-3}{3}=\frac{y+5}{-7}=\frac{z-1}{-10}\end{array}$

Therefore, the set of symmetric equation is $\frac{x-3}{3}=\frac{y+5}{-7}=\frac{z-1}{-10}$ .