Chapter 10.4, Problem 15E

(a)

To determine

To find:The set of parametric equations.

Answer to Problem 15E

Theparametric equations are $x=-3+4t$ , $y=8-10t$ and $z=15+t$ .

Explanation of Solution

Given information:The points are $\left(-3,8,15\right)$ and $\left(1,-2,16\right)$ .

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:

  $\begin{array}{c}v=\langle 1+3,-2-8,16-15\rangle \\ =\langle 4,-10,1\rangle \end{array}$

So, the normal vector is $v=\langle 4,-10,1\rangle$ and point on the plane is $\left(-3,8,15\right)$ .

Substitute $-3$ for $x_1$ and 4 for $a$ in the equation $x=x_1+at$ .

  $x=-3+4t$

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

  $y=8-10t$

Substitute 15 for $z_1$ and 1 for $c$ in the equation $z=z_1+ct$ .

  $z=15+t$

Therefore, the parametric equations are $x=-3+4t$ , $y=8-10t$ and $z=15+t$ .

(b)

To determine

To find:The set of symmetric equations for the line passes through the two points $\left(-3,8,15\right)$ and $\left(1,-2,16\right)$ .

Answer to Problem 15E

Theset of symmetric equation is $\frac{x+3}{4}=\frac{y-8}{-10}=\frac{z-15}{1}$ .

Explanation of Solution

Given information: The points are $\left(-3,8,15\right)$ and $\left(1,-2,16\right)$ .

Calculation:

The normal vector is $v=\langle 4,-10,1\rangle$ . It can be seen that the direction numbers are all non-zero.So, the set of symmetric equations can be calculated as:

  $\begin{array}{c}\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}\\ \frac{x+3}{4}=\frac{y-8}{-10}=\frac{z-15}{1}\end{array}$

Therefore, the set of symmetric equation is $\frac{x+3}{4}=\frac{y-8}{-10}=\frac{z-15}{1}$ .