Chapter 11.4, Problem 7E

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 7E

$x=-4+6t,y=1+8t,z=-6t$

Explanation of Solution

Given information:

A line is passing through point $(-4,1,0)$ and is parallel to $\text{v}=\frac{1}{2}i+\frac{4}{3}j-k$ .

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 $(-4,1,0)$ .

So, $x_1=-4,y_1=1,z_1=0$ .

Line is parallel to the direction vector $\text{v}=\frac{1}{2}i+\frac{4}{3}j-k$ .

$\Rightarrow \text{v}=6i+8j-6k$$\left\{\text{Converting direction number into integer}\right\}$

So, direction numbers are $a=6,b=8,c=-6$ .

Now parametric equation of the lines passing through point $(-4,1,0)$ and is parallel to $\text{v}=6i+8j-6k$ will be given as-

$\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=-4+6t,y=1+8t,z=0-6t\\ \Rightarrow x=-4+6t,y=1+8t,z=-6t\end{array}$

Hence, desired parametric equation of line is $x=-4+6t,y=1+8t,z=-6t$ .

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 7E

$\frac{x+4}{6}=\frac{y-1}{8}=-\frac{z}{6}$ .

Explanation of Solution

Given information:

A line is passing through point $(-4,1,0)$ and is parallel to $\text{v}=\frac{1}{2}i+\frac{4}{3}j-k$ .

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 $(-4,1,0)$ .

So, $x_1=-4,y_1=1,z_1=0$ .

Line is parallel to the direction vector $\text{v}=\frac{1}{2}i+\frac{4}{3}j-k$ .

$\Rightarrow \text{v}=6i+8j-6k$$\left\{\text{Converting direction number into integer}\right\}$

So, direction numbers are $a=6,b=8,c=-6$ .

From part $a$ of this exercise, parametric equation of the lines passing through point $(-4,1,0)$ and is parallel to $\text{v}=6i+8j-6k$ is given by -

$x=-4+6t,y=1+8t,z=-6t$ .

Since, direction numbers $a=6,b=8,c=-6$ are all non-zero.

Therefore, set of symmetric equation will be given as −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-(-4)}{6}=\frac{y-1}{8}=\frac{z-0}{-6}\\ \Rightarrow \frac{x+4}{6}=\frac{y-1}{8}=-\frac{z}{6}\end{array}$

Hence, desired set of symmetric equation of line is $\frac{x+4}{6}=\frac{y-1}{8}=-\frac{z}{6}$ .