Chapter 11.4, Problem 10E

a.

To determine

ToFind:A set of parametric equations of the line passing through the point $\left(4,1,2\right)$ and parallel to the vector $v=\langle 5,3,6\rangle$

Answer to Problem 10E

The set of parametric equations of the line passing through the given point and parallel to the given vector is

  $x=4+5t$ ,   $y=1+3t$ and   $z=2+6t$

Explanation of Solution

Given:

A line passing through the point $\left(4,1,2\right)$ and parallel to the vector $v=\langle 5,3,6\rangle$

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 point $\left(4,1,2\right)$ and parallel to the vector $v=\langle 5,3,6\rangle$

Using the coordinates of the point

  $x_1=4$ , $y_1=1$ and $z_1=2$

And the direction numbers as

  $a=5$ , $b=3$ and $c=6$

The set of parametric equations of the line is

  $x=4+5t$ ,   $y=1+3t$ and   $z=2+6t$

Conclusion:

The set of parametric equations of the line passing through the given point and parallel to the given vector is

  $x=4+5t$ ,   $y=1+3t$ and   $z=2+6t$

b.

To determine

ToFind:A set of symmetric equations of the line passing through the point $\left(4,1,2\right)$ and parallel to the vector $v=\langle 5,3,6\rangle$

Answer to Problem 10E

The set of symmetric equations of the line passing through the given point and parallel to the given vector is

  $\frac{x-4}{5}=\frac{y-1}{3}=\frac{z-2}{6}$

Explanation of Solution

Given:

A line passing through the point $\left(4,1,2\right)$ and parallel to the vector $v=\langle 5,3,6\rangle$

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 point $\left(4,1,2\right)$ and parallel to the vector $v=\langle 5,3,6\rangle$

Using the coordinates of the point

  $x_1=4$ , $y_1=1$ and $z_1=2$

And the direction numbers as

  $a=5$ , $b=3$ and $c=6$

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

  $\frac{x-4}{5}=\frac{y-1}{3}=\frac{z-2}{6}$

Conclusion:

The set of symmetric equations of the line passing through the given point and parallel to the given vector is

  $\frac{x-4}{5}=\frac{y-1}{3}=\frac{z-2}{6}$