Chapter 11, Problem 61RE

a.

To determine

ToFind:A set of parametric equations of the line passing through the points $\left(-1,3,5\right)$ and $\left(3,6,-1\right)$

Answer to Problem 61RE

The set of parametric equations of the line passing through the given points is

  $x=-1+4t$ ,   $y=3+3t$ and   $z=5-6t$

Explanation of Solution

Given:

A line passing through the points $\left(-1,3,5\right)$ and $\left(3,6,-1\right)$

Concept Used:

A line $L$ passing through the point $P=\left(x_1,y_1,z_1\right)$ and parallel to the vector $v=\langle a,b,c\rangle$ is represented by the parametric equations

  $x=x_1+at$ , $y=y_1+bt$ and $z=z_1+ct$

Calculation:

Forthe line passing through the points $\left(-1,3,5\right)$ and $\left(3,6,-1\right)$

Let $P=\left(-1,3,5\right)$ and $Q=\left(3,6,-1\right)$

Then $P=\left(-1,3,5\right)$ a direction vector for the line passing through $P$ and $Q$ is

  $\begin{array}{c}v=\overrightarrow{PQ}\\ =\langle 3+1,6-3,-1-5\rangle \\ =\langle 4,3,-6\rangle \end{array}$

Using the coordinates of the initial point $P=\left(-1,3,5\right)$

  $x_1=-1$ , $y_1=3$ and $z_1=5$

And the direction numbers as

  $a=4$ , $b=3$ and $c=-6$

The set of parametric equations of the line is

  $x=-1+4t$ ,   $y=3+3t$ and   $z=5-6t$

Conclusion:

The set of parametric equations of the line passing through the given points is

  $x=-1+4t$ ,   $y=3+3t$ and   $z=5-6t$

b.

To determine

ToFind: A set of symmetric equations of the line passing through the points $\left(-1,3,5\right)$ and $\left(3,6,-1\right)$

Answer to Problem 61RE

The set of symmetric equations of the line passing through the given points is

  $\frac{x+1}{4}=\frac{y-3}{3}=\frac{z-5}{-6}$

Explanation of Solution

Given:

A line passing through the points $\left(-1,3,5\right)$ and $\left(3,6,-1\right)$

Concept Used:

A line $L$ passing through the point $P=\left(x_1,y_1,z_1\right)$ and parallel to the vector $v=\langle a,b,c\rangle$ is represented by the parametric equations

  $x=x_1+at$ , $y=y_1+bt$ and $z=z_1+ct$

When the direction numbers $a$ , $b$ and $c$ are all nonzero, then symmetric equations of the line is given by

  $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Calculation:

For the line passing through the points $\left(-1,3,5\right)$ and $\left(3,6,-1\right)$

Let $P=\left(-1,3,5\right)$ and $Q=\left(3,6,-1\right)$

Then $P=\left(-1,3,5\right)$ a direction vector for the line passing through $P$ and $Q$ is

  $\begin{array}{c}v=\overrightarrow{PQ}\\ =\langle 3+1,6-3,-1-5\rangle \\ =\langle 4,3,-6\rangle \end{array}$

Using the coordinates of the initial point $P=\left(-1,3,5\right)$

  $x_1=-1$ , $y_1=3$ and $z_1=5$

And the direction numbers as

  $a=4$ , $b=3$ and $c=-6$

Since, the direction numbers $a$ , $b$ and $c$ are all nonzero, the symmetric equations of the line is given by

  $\frac{x+1}{4}=\frac{y-3}{3}=\frac{z-5}{-6}$

Conclusion:

The set of symmetric equations of the line passing through the given points is

  $\frac{x+1}{4}=\frac{y-3}{3}=\frac{z-5}{-6}$