Chapter 11, Problem 60RE

a.

To determine

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

Answer to Problem 60RE

  $x=5t,y=-10+20t,z=3-3t$

Explanation of Solution

Given information:

A line is passing through point $(0,-10,3)$ and $(5,10,0)$ .

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 $P$

$(0,-10,3)$ and $Q$

$(5,10,0)$ .

So, $x_1=0,y_1=-10,z_1=3$ .

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

$\begin{array}{l}\text{v}=\overrightarrow{PQ}=\langle 5-0,10-(-10),0-3\rangle \\ \text{v}=\overrightarrow{PQ}=\langle 5,20,-3\rangle \end{array}$

Line is parallel to the direction vector $\text{v}=\overrightarrow{PQ}=\langle 5,20,-3\rangle$ .

So, direction numbers are $a=5,b=20,c=-3$ .

Now parametric equation of the lines passing through point $(0,-10,3)$ and is parallel to $\text{v}=\overrightarrow{PQ}=\langle 5,20,-3\rangle$ will be given as-

  $\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=0+5t,y=-10+20t,z=3-3t\\ \Rightarrow x=5t,y=-10+20t,z=3-3t\end{array}$

Hence, desired parametric equation of line is $x=5t,y=-10+20t,z=3-3t$ .

b.

To determine

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

Answer to Problem 60RE

$\frac{x}{5}=\frac{y+10}{20}=-\frac{z-3}{3}$ .

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 $P$

$(0,-10,3)$ and $Q$

$(5,10,0)$ .

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

  $x=5t,y=-10+20t,z=3-3t$ .

Since, direction numbers $a=5,b=20,c=-3$ are all non-zero.

Therefore, set of symmetric equation will be given as −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-0}{5}=\frac{y-(-10)}{20}=\frac{z-3}{-3}\\ \Rightarrow \frac{x}{5}=\frac{y+10}{20}=-\frac{z-3}{3}\end{array}$

Hence, desired set of symmetric equation of line is $\frac{x}{5}=\frac{y+10}{20}=-\frac{z-3}{3}$ .