Step1

Given:

The vector-valued function \(\displaystyle{r}{\left({t}\right)}={\left({t}+{1}\right)}{i}+{\left({3}{t}-{1}\right)}{j}+{2}{t}{k}\) and it gives the orientation of the curve.

Step2

The vector-valued function is given below as,

\(\displaystyle{r}{\left({t}\right)}={\left({t}+{1}\right)}{i}+{\left({3}{t}-{1}\right)}{j}+{2}{t}{k}\)

The first two parametric equations is given below as,

\(\displaystyle{x}={t}+{i}{\quad\text{and}\quad}{y}={3}{t}-{1}\)

The rectangular equation determines the two parametric equations.

The value of t from x=t+ 1 is calculated as below,

\(\displaystyle{x}={t}+{i}\)

\(\displaystyle{x}-{1}={t}+\cancel{{{1}}}-\cancel{{{1}}}\)

\(\displaystyle{t}={x}-{1}\)

The value of t=x-1.

Step3

On Substituting the value of t in y= 3t -1 to get

\(\displaystyle{y}={3}{\left({x}-{1}\right)}-{1}\)

\(\displaystyle={3}{x}-{3}-{1}\)

\(\displaystyle={3}{x}-{4}\)

Hence, the value of y=3x-4.

Thus, the rectangular equation is \(\displaystyle{y}={3}{x}-{4}\)

Hence, the curve is a straight line that is determined with respect to the space.

The curve move upwards as the value of t increases resulting in the third parametric equation that is z= 2t.

The vector valued function of the graph is being represented and the orientation is shown in the graph as,

\(\displaystyle{r}{\left({t}\right)}={\left({t}+{1}\right)}{i}+{\left({3}{t}-{1}\right)}{j}+{2}{t}{k}\)

Given:

The vector-valued function \(\displaystyle{r}{\left({t}\right)}={\left({t}+{1}\right)}{i}+{\left({3}{t}-{1}\right)}{j}+{2}{t}{k}\) and it gives the orientation of the curve.

Step2

The vector-valued function is given below as,

\(\displaystyle{r}{\left({t}\right)}={\left({t}+{1}\right)}{i}+{\left({3}{t}-{1}\right)}{j}+{2}{t}{k}\)

The first two parametric equations is given below as,

\(\displaystyle{x}={t}+{i}{\quad\text{and}\quad}{y}={3}{t}-{1}\)

The rectangular equation determines the two parametric equations.

The value of t from x=t+ 1 is calculated as below,

\(\displaystyle{x}={t}+{i}\)

\(\displaystyle{x}-{1}={t}+\cancel{{{1}}}-\cancel{{{1}}}\)

\(\displaystyle{t}={x}-{1}\)

The value of t=x-1.

Step3

On Substituting the value of t in y= 3t -1 to get

\(\displaystyle{y}={3}{\left({x}-{1}\right)}-{1}\)

\(\displaystyle={3}{x}-{3}-{1}\)

\(\displaystyle={3}{x}-{4}\)

Hence, the value of y=3x-4.

Thus, the rectangular equation is \(\displaystyle{y}={3}{x}-{4}\)

Hence, the curve is a straight line that is determined with respect to the space.

The curve move upwards as the value of t increases resulting in the third parametric equation that is z= 2t.

The vector valued function of the graph is being represented and the orientation is shown in the graph as,

\(\displaystyle{r}{\left({t}\right)}={\left({t}+{1}\right)}{i}+{\left({3}{t}-{1}\right)}{j}+{2}{t}{k}\)