Chapter 8.6, Problem 36E

Solution

To find: The vector and parametric forms of the line through the point $P_0$ in the direction of $\text{v}$ .

The vector form of the line through the point $P_0\left(-3,8,-1\right)$ in the direction of $\text{v=}〈-3,5,2〉$ is $r=〈-3,8,-1〉+t〈-3,5,2〉$ .

The parametric form of the line through the point $P_0\left(-3,8,-1\right)$ in the direction of $\text{v=}〈-3,5,2〉$ are . $x=-3-3t$ , $y=8+5t$ , $z=-1+2t$ .

Given information:

The given point $P_0\left(-3,8,-1\right)$ and vector $\text{v=}〈-3,5,2〉$ .

Formula used:

If $l$ be a line through point $P_0\left(x_0,y_0,z_0\right)$ in the direction of a non-zero vector $\text{v=}〈a,b,c〉$ and a point $P\left(x,y,z\right)$ is on the line then-

Vector form of line:- $\text{r}={\text{r}}_{\text{0}}+t\text{v}$ , where $\text{r}=〈x,y,z〉$ and $r_0=〈x_0,y_0,z_0〉$ .

Parametric form of line:- $x=x_0+at$ , $y=y_0+bt$ and $z=z_0+ct$ .

Where $t$ is a real number.

Calculation:

Consider the point $P_0\left(-3,8,-1\right)$ and vector $\text{v=}〈-3,5,2〉$ .

Using the formula $\text{r}={\text{r}}_{\text{0}}+t\text{v}$

Substitute $〈-3,8,-1〉$ for ${\text{r}}_0$ and $〈-3,5,2〉$ for $\text{v}$ .

  $r=〈-3,8,-1〉+t〈-3,5,2〉$

Using the formula $x=x_0+at$ , $y=y_0+bt$ and $z=z_0+ct$

Substitute $-3$ for $x_0$ , $8$ for $y_0$ , $-1$ for $z_0$ , $-3$ for $a$ , $5$ for $b$ , $2$ for $c$

  $\begin{array}{c}x=-3-3t\\ y=8+5t\\ z=-1+2t\end{array}$

Thus the vector form of the line through the point $P_0\left(-3,8,-1\right)$ in the direction of $\text{v=}〈-3,5,2〉$ is $r=〈-3,8,-1〉+t〈-3,5,2〉$ .

And the parametric form of the line through the point $P_0\left(-3,8,-1\right)$ in the direction of $\text{v=}〈-3,5,2〉$ are $x=-3-3t$ , $y=8+5t$ , $z=-1+2t$ .