Chapter 11.4, Problem 18E

a.

To determine

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

Answer to Problem 18E

  $x=-\frac{3}{2}+9t,y=\frac{3}{2}-13t,z=2-4t$

Explanation of Solution

Given information:

A line is passing through point $\left(-\frac{3}{2},\frac{3}{2},2\right)$ and $\left(3,-5,4\right)$ .

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 $P$

$\left(-\frac{3}{2},\frac{3}{2},2\right)$ and $Q$

$\left(3,-5,4\right)$ .

So, $x_1=-\frac{3}{2},y_1=\frac{3}{2},z_1=2$ .

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

$\text{v}=\overrightarrow{PQ}=\langle 3-\left(-\frac{3}{2}\right),-5-\frac{3}{2},4-2\rangle$

$\begin{array}{l}\text{v}=\overrightarrow{PQ}=\langle \frac{9}{2},-\frac{13}{2},2\rangle \\ \Rightarrow \text{v}=\overrightarrow{PQ}=\langle 9,-13,4\rangle \end{array}$

Line is parallel to the direction vector $\text{v}=\overrightarrow{PQ}=\langle 9,-13,4\rangle$ .

So, direction numbers are $a=9,b=-13,c=4$ .

Now parametric equation of the lines passing through point $\left(-\frac{3}{2},\frac{3}{2},2\right)$ and is parallel to $\text{v}=\overrightarrow{PQ}=\langle 9,-13,4\rangle$ will be given as-

  $\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=-\frac{3}{2}+9t,y=\frac{3}{2}-13t,z=2-4t\end{array}$

Hence, desired parametric equation of line is $x=-\frac{3}{2}+9t,y=\frac{3}{2}-13t,z=2-4t$ .

b.

To determine

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

Answer to Problem 18E

$\frac{2x+3}{18}=-\frac{2y-3}{26}=-\frac{z-2}{4}$

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:

A line is passing through point $\left(-\frac{3}{2},\frac{3}{2},2\right)$ and $\left(3,-5,4\right)$ .

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

  $x=-\frac{3}{2}+9t,y=\frac{3}{2}-13t,z=2-4t$ .

Since, direction numbers $a=9,b=-13,c=4$ are all non-zero.

Therefore, set of symmetric equation will be given by −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-\left(-\frac{3}{2}\right)}{9}=\frac{y-\frac{3}{2}}{-13}=\frac{z-2}{4}\\ \Rightarrow \frac{2x+3}{18}=-\frac{2y-3}{26}=-\frac{z-2}{4}\end{array}$

Hence, desired set of symmetric equation is $\frac{2x+3}{18}=-\frac{2y-3}{26}=-\frac{z-2}{4}$ .