Chapter 10.4, Problem 17E

(a)

To determine

To find:The set of parametric equations.

Answer to Problem 17E

Theset of parametric equations are $x=-\frac{1}{2}+\frac{3}{2}t$ , $y=2-\frac{5}{2}t$ and $z=\frac{1}{2}-\frac{1}{2}t$ .

Explanation of Solution

Given information:The points are $\left(-\frac{1}{2},2,\frac{1}{2}\right)$ and $\left(1,-\frac{1}{2},0\right)$ .

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:

  $\begin{array}{c}v=\langle 1+\frac{1}{2},-\frac{1}{2}-2,0-\frac{1}{2}\rangle \\ =\langle \frac{3}{2},-\frac{5}{2},-\frac{1}{2}\rangle \end{array}$

So, the normal vector is $v=\langle \frac{3}{2},-\frac{5}{2},-\frac{1}{2}\rangle$ and point on the plane is $\left(-\frac{1}{2},2,\frac{1}{2}\right)$ .

Substitute $-\frac{1}{2}$ for $x_1$ and $\frac{3}{2}$ for $a$ in the equation $x=x_1+at$ .

  $x=-\frac{1}{2}+\frac{3}{2}t$

Substitute 2 for $y_1$ and $-\frac{5}{2}$ for $b$ in the equation $y=y_1+bt$ .

  $y=2-\frac{5}{2}t$

Substitute $\frac{1}{2}$ for $z_1$ and $-\frac{1}{2}$ for $c$ in the equation $z=z_1+ct$ .

  $z=\frac{1}{2}-\frac{1}{2}t$

Therefore, the set of parametric equations are $x=-\frac{1}{2}+\frac{3}{2}t$ , $y=2-\frac{5}{2}t$ and $z=\frac{1}{2}-\frac{1}{2}t$ .

(b)

To determine

To find:The set of symmetric equations for the line through two points $\left(-\frac{1}{2},2,\frac{1}{2}\right)$ and $\left(1,-\frac{1}{2},0\right)$ .

Answer to Problem 17E

Theset of symmetric equation is $\frac{x+1/2}{3/2}=\frac{y-2}{-5/2}=\frac{z-1/2}{-1/2}$ .

Explanation of Solution

Given information:The points are $\left(-\frac{1}{2},2,\frac{1}{2}\right)$ and $\left(1,-\frac{1}{2},0\right)$ .

Calculation:

The normal vector is $v=\langle \frac{3}{2},-\frac{5}{2},-\frac{1}{2}\rangle$ and point on the plane is $\left(-\frac{1}{2},2,\frac{1}{2}\right)$ .It can be seen that the direction numbers are all non-zero. So, the set of symmetric equations can be calculated as:

  $\begin{array}{c}\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}\\ \frac{x+1/2}{3/2}=\frac{y-2}{-5/2}=\frac{z-1/2}{-1/2}\end{array}$

Therefore, the set of symmetric equation is $\frac{x+1/2}{3/2}=\frac{y-2}{-5/2}=\frac{z-1/2}{-1/2}$ .