To find the vector equation of the line that is parallel to the [tex]\( z \)[/tex]-axis and passes through the point [tex]\((2, -3, 5)\)[/tex], follow these steps:
1. Identify the given point:
The line passes through the point [tex]\((2, -3, 5)\)[/tex]. This point can be written in vector form as:
[tex]\[
\vec{r_0} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}
\][/tex]
2. Determine the direction vector:
Since the line is parallel to the [tex]\( z \)[/tex]-axis, its direction vector will have a nonzero component only in the [tex]\( z \)[/tex]-direction. Hence, the direction vector can be written as:
[tex]\[
\vec{d} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\][/tex]
3. Formulate the vector equation of the line:
The general vector equation of a line that passes through a point [tex]\(\vec{r_0}\)[/tex] and has a direction vector [tex]\(\vec{d}\)[/tex] is:
[tex]\[
\vec{r}(t) = \vec{r_0} + t \vec{d}
\][/tex]
where [tex]\( t \)[/tex] is a parameter.
4. Substitute the given point and direction vector:
[tex]\[
\vec{r}(t) = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + t \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\][/tex]
5. Simplify the equation:
Adding the vectors together, we get:
[tex]\[
\vec{r}(t) = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ t \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 + t \end{pmatrix}
\][/tex]
Therefore, the vector equation of the line parallel to the [tex]\( z \)[/tex]-axis and passing through the point [tex]\((2, -3, 5)\)[/tex] is:
[tex]\[
\vec{r}(t) = \begin{pmatrix} 2 \\ -3 \\ 5 + t \end{pmatrix}
\][/tex]
