Chapter 11.4, Problem 17E

a.

To determine

To find: A set of parametric equation of the lines passing through given two points.

Answer to Problem 17E

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

Explanation of Solution

Given information:

A line is passing through point $\left(-\frac{1}{2},2,\frac{1}{2}\right)$ and $\left(1,-\frac{1}{2},0\right)$ .

Concepts used:

Vectors are used for determining equation of a line in space. A line passing through the point $P(x_1,y_1,z_1)$ and parallel to the vector $\text{v}=\langle a,b,c\rangle$ is represented by the Parametric equations-

$x=x_1+at,y=y_1+bt,z=z_1+ct$ ;where, $a,b,c$ are direction numbers of vector $\text{v}=\langle a,b,c\rangle$ .

Calculation:

Let line is passing through the points $P$

$\left(-\frac{1}{2},2,\frac{1}{2}\right)$ and $Q$

$\left(1,-\frac{1}{2},0\right)$ .

So, $x_1=-\frac{1}{2},y_1=1,z_1=\frac{1}{2}$ .

Also, the direction vector of line passing through $P$ and $Q$ is

$\text{v}=\overrightarrow{PQ}=\langle 1-\left(-\frac{1}{2}\right),-\frac{1}{2}-2,0-\frac{1}{2}\rangle$

$\begin{array}{l}\text{v}=\overrightarrow{PQ}=\langle \frac{3}{2},-\frac{5}{2},-\frac{1}{2}\rangle \\ \Rightarrow \text{v}=\overrightarrow{PQ}=\langle 3,-5,-1\rangle \end{array}$

Line is parallel to the direction vector $\text{v}=\overrightarrow{PQ}=\langle 3,-5,-1\rangle$ .

So, direction numbers are $a=3,b=-5,c=-1$ .

Now parametric equation of the lines passing through point $\left(-\frac{1}{2},2,\frac{1}{2}\right)$ and is parallel to $\text{v}=\overrightarrow{PQ}=\langle 3,-5,-1\rangle$ will be given as-

  $\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=-\frac{1}{2}+3t,y=2-5t,z=\frac{1}{2}-t\end{array}$

Hence, desired parametric equation of line is $x=-\frac{1}{2}+3t,y=2-5t,z=\frac{1}{2}-t$ .

b.

To determine

To find: A set of symmetric equation of the line passing through two given points.

Answer to Problem 17E

$\frac{2x+1}{6}=-\frac{y-2}{5}=-\frac{2z-1}{2}$

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Concepts used:

Vectors are used for determining equation of a line in space. A line passing through the point $P(x_1,y_1,z_1)$ and parallel to the vector $\text{v}=\langle a,b,c\rangle$ is represented by the Parametric equations-

$x=x_1+at,y=y_1+bt,z=z_1+ct$ ;where, $a,b,c$ are direction numbers of vector $\text{v}=\langle a,b,c\rangle$ .

When direction numbers $a,b,c$ are non-zero then parameter $t$ can be eliminated to obtain set of symmetric equation $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Calculation:

Let line is passing through the points $P$

$\left(-\frac{1}{2},2,\frac{1}{2}\right)$ and $Q$

$\left(1,-\frac{1}{2},0\right)$ .

From part $a$ of this exercise, parametric equation of the line is given by -

  $x=-\frac{1}{2}+3t,y=2-5t,z=\frac{1}{2}-t$ .

Since, direction numbers $a=3,b=-5,c=-1$ are all non-zero.

Therefore, set of symmetric equation will be given by −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-\left(-\frac{1}{2}\right)}{3}=\frac{y-2}{-5}=\frac{z-\left(\frac{1}{2}\right)}{-1}\\ \Rightarrow \frac{2x+1}{6}=-\frac{y-2}{5}=-\frac{2z-1}{2}\end{array}$

Hence, desired set of symmetric equation is $\frac{2x+1}{6}=-\frac{y-2}{5}=-\frac{2z-1}{2}$ .