Chapter 11.4, Problem 45E

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 45E

  $x=5+2t,y=-3-t,z=-4+3t$

Explanation of Solution

Given information:

A line is passing through point $(5,-3,-4)$ and is parallel to $\text{v}=\langle 2,-1,3\rangle$ .

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 $(5,-3,-4)$ .

So, $x_1=5,y_1=-3,z_1=-4$ .

Line is parallel to the direction vector $\text{v}=\langle 2,-1,3\rangle$ .

So, direction numbers are $a=2,b=-1,c=3$ .

Now parametric equation of the lines passing through point $(5,-3,-4)$ and is parallel to $\text{v}=\langle 2,-1,3\rangle$ will be given as-

  $\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=5+2t,y=-3-t,z=-4+3t\end{array}$

Hence, desired parametric equation of line is $x=5+2t,y=-3-t,z=-4+3t$ .