Chapter 8.6, Problem 41E

Solution

To find: The vector equation of line through $A$ and $B$ .

The vector equation of line through $A$ and $B$ is $\text{r}=〈-1,2,4〉+t〈1,4,-7〉$ .

Given information:

The given points are $A\left(-1,2,4\right)$ and $B\left(0,6,-3\right)$ .

Formula used:

If $A=\left(a_1,a_2,a_3\right)$ and $B=\left(b_1,b_2,b_3\right)$ then vector $\overrightarrow{AB}=\left(b_1-a_1,b_2-a_2,b_3-a_3\right)$ .

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 $r_0=〈x_0,y_0,z_0〉$ .

where $t$ is a real number.

Calculation:

Consider the point $A\left(-1,2,4\right)$ and $B\left(0,6,-3\right)$ .

The line is in the direction of $\text{v}=\overrightarrow{AB}$

Using the formula $\overrightarrow{AB}=\left(b_1-a_1,b_2-a_2,b_3-a_3\right)$

Substitute $-1$ for $a_1$ , $2$ for $a_2$ , $4$ for $a_3$ , $0$ for $b_1$ , $6$ for $b_2$ and $-3$ for $b_3$ .

  $\begin{array}{c}\overrightarrow{AB}=\left(0-\left(-1\right),6-2,-3-4\right)\\ \text{v}=\left(1,4,-7\right)\end{array}$

Vector form of line Using the formula $\text{r}={\text{r}}_{\text{0}}+t\text{v}$

Substitute $〈-1,2,4〉$ for ${\text{r}}_0$ and $〈1,4,-7〉$ for $\text{v}$ .

  $\text{r}=〈-1,2,4〉+t〈1,4,-7〉$

Thus, the vector equation of the line through $A$ and $B$ is $\text{r}=〈-1,2,4〉+t〈1,4,-7〉$ .