Chapter 11.4, Problem 8E

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 8E

$x=-2+2t,y=4t,z=3-2t$

Explanation of Solution

Given information:

A line is passing through point $(-2,0,3)$ and is parallel to $\text{v}=2i+4j-2k$ .

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 $(-2,0,3)$ .

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

Line is parallel to the direction vector $\text{v}=2i+4j-2k$ .

So, direction numbers are $a=2,b=4,c=-2$ .

Now parametric equation of the lines passing through point $(-2,0,3)$ and is parallel to $\text{v}=2i+4j-2k$ will be given as-

$\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=-2+2t,y=0+4t,z=3-2t\\ \Rightarrow x=-2+2t,y=4t,z=3-2t\end{array}$

Hence, desired parametric equation of line is $x=-2+2t,y=4t,z=3-2t$ .

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 8E

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

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:

Line is passing through the point $(-2,0,3)$ .

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

Line is parallel to the direction vector $\text{v}=2i+4j-2k$ .

So, direction numbers are $a=2,b=4,c=-2$ .

From part $a$ of this exercise, parametric equation of the lines passing through point $(-2,0,3)$ and is parallel to $\text{v}=2i+4j-2k$ is given by -

$x=-2+2t,y=4t,z=3-2t$ .

Since, direction numbers $a=2,b=4,c=-2$ are all non-zero.

Therefore, set of symmetric equation will be given as −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-(-2)}{2}=\frac{y-0}{4}=\frac{z-3}{-2}\\ \Rightarrow \frac{x+2}{2}=\frac{y}{4}=-\frac{z-3}{2}\end{array}$

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