Chapter 11.4, Problem 13E

a.

To determine

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

Answer to Problem 13E

  $x=-3+4t,y=8-10t,z=15+t$

Explanation of Solution

Given information:

A line is passing through point $(-3,8,15)$ and $(1,-2,16)$ .

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 $P$

$(-3,8,15)$ and $Q$

$(1,-2,16)$ .

So, $x_1=-3,y_1=8,z_1=15$ .

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

$\text{v}=\overrightarrow{PQ}=\langle 1-(-3),-2-8,16-15\rangle$

$\text{v}=\overrightarrow{PQ}=\langle 4,-10,1\rangle$

Line is parallel to the direction vector $\text{v}=\overrightarrow{PQ}=\langle 4,-10,1\rangle$ .

So, direction numbers are $a=4,b=-10,c=1$ .

Now parametric equation of the lines passing through point $(-3,8,15)$ and is parallel to $\text{v}=\overrightarrow{PQ}=\langle 4,-10,1\rangle$ will be given as-

  $\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=-3+4t,y=8-10t,z=15+t\\ \Rightarrow x=-3+4t,y=8-10t,z=15+t\end{array}$

Hence, desired parametric equation of line is $x=-3+4t,y=8-10t,z=15+t$ .

b.

To determine

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

Answer to Problem 13E

$\frac{x+3}{4}=-\frac{y-8}{10}=\frac{z-15}{1}$ .

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 $P$

$(-3,8,15)$ and $Q$

$(1,-2,16)$ .

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

  $x=-3+4t,y=8-10t,z=15+t$ .

Since, direction numbers $a=4,b=-10,c=1$ are all non-zero.

Therefore, set of symmetric equation will be given as −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-(-3)}{4}=\frac{y-8}{-10}=\frac{z-15}{1}\\ \Rightarrow \frac{x+3}{4}=-\frac{y-8}{10}=\frac{z-15}{1}\end{array}$

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