Chapter 11.4, Problem 12E

a.

To determine

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

Answer to Problem 12E

  $x=2+8t,y=3+5t,z=12t$

Explanation of Solution

Given information:

A line is passing through point $(2,3,0)$ and $(10,8,12)$ .

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 $P$

$(2,3,0)$ and $Q$

$(10,8,12)$ .

So, $x_1=2,y_1=3,z_1=0$ .

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

$\text{v}=\overrightarrow{PQ}=\langle 10-2,8-3,12-0\rangle$

$\text{v}=\overrightarrow{PQ}=\langle 8,5,12\rangle$

Line is parallel to the direction vector $\text{v}=\overrightarrow{PQ}=\langle 8,5,12\rangle$ .

So, direction numbers are $a=8,b=5,c=12$ .

Now parametric equation of the lines passing through point $(2,3,0)$ and is parallel to $\text{v}=\overrightarrow{PQ}=\langle 8,5,12\rangle$ will be given as-

  $\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=2+8t,y=3+5t,z=0+12t\\ \Rightarrow x=2+8t,y=3+5t,z=12t\end{array}$

Hence, desired parametric equation of line is $x=2+8t,y=3+5t,z=12t$ .

b.

To determine

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

Answer to Problem 12E

$\frac{x-2}{8}=\frac{y-3}{5}=\frac{z}{12}$ .

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 $P$

$(2,3,0)$ and $Q$

$(10,8,12)$ .

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

  $x=2+8t,y=3+5t,z=12t$ .

Since, direction numbers $a=8,b=5,c=12$ are all non-zero.

Therefore, set of symmetric equation will be given as −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-2}{8}=\frac{y-3}{5}=\frac{z-0}{12}\\ \Rightarrow \frac{x-2}{8}=\frac{y-3}{5}=\frac{z}{12}\end{array}$

Hence, desired set of symmetric equation of line is $\frac{x-2}{8}=\frac{y-3}{5}=\frac{z}{12}$ .