Chapter 11.4, Problem 6E

a.

To determine

To find: A set of parametric equation of the lines passing through given point and parallel to the specified vector.

Answer to Problem 6E

$x=3+3t,y=-5-7t,z=1-10t$

Explanation of Solution

Given information:

A line is passing through point $(3,-5,1)$ and is parallel to $\text{v}=\langle 3,-7,-10\rangle$ .

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:

Line is passing through the point $(3,-5,1)$ .

So, $x_1=3,y_1=-5,z_1=1$ .

Line is parallel to the direction vector $\text{v}=\langle 3,-7,-10\rangle$ .

So, direction numbers are $a=3,b=-7,c=-10$ .

Now parametric equation of the lines passing through point $(3,-5,1)$ and is parallel to $\text{v}=\langle 3,-7,-10\rangle$ will be given as-

$\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=3+3t,y=-5-7t,z=1-10t\end{array}$

Hence, desired parametric equation of line is $x=3+3t,y=-5-7t,z=1-10t$ .

b.

To determine

To find: A set of symmetric equation of the line passing through given point and parallel to the specified vector.

Answer to Problem 6E

$\frac{x-3}{3}=-\frac{y+5}{7}=-\frac{z-1}{10}$ .

Explanation of Solution

Given information:

A line is passing through point $(3,-5,1)$ and is parallel to $\text{v}=\langle 3,-7,-10\rangle$ .

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:

Line is passing through the point $(3,-5,1)$ .

So, $x_1=3,y_1=-5,z_1=1$ .

Line is parallel to the direction vector $\text{v}=\langle 3,-7,-10\rangle$ .

So, direction numbers are $a=3,b=-7,c=-10$ .

From part $a$ of this exercise, parametric equation of the lines passing through point $(3,-5,1)$ and is parallel to $\text{v}=\langle 3,-7,-10\rangle$ is given by -

$x=3+3t,y=-5-7t,z=1-10t$ .

Since, direction numbers $a=3,b=-7,c=-10$ are all non-zero.

Therefore, set of symmetric equation will be given as −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-3}{3}=\frac{y-(-5)}{-7}=\frac{z-1}{-10}\\ \Rightarrow \frac{x-3}{3}=-\frac{y+5}{7}=-\frac{z-1}{10}\end{array}$

Hence, desired set of symmetric equation of line is $\frac{x-3}{3}=-\frac{y+5}{7}=-\frac{z-1}{10}$ .