Chapter 8.6, Problem 43E

Solution

To find: The parametric equations for the line through $A$ and $C$ .

The parametric equations for the line through $A$ and $C$ are $x=-1+3t$ , $y=2-6t$ , and $z=4-3t$ .

Given information:

The given points are $A\left(-1,2,4\right)$ and $C\left(2,-4,1\right)$ .

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-

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 $A\left(-1,2,4\right)$ and $C\left(2,-4,1\right)$ .

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

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 $a_3$ , $2$ for $b_1$ , $-4$ for $b_2$ and $1$ for $b_3$ .

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

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

Substitute $-1$ for $x_0$ , $2$ for $y_0$ , $4$ for $z_0$ , $3$ for $a$ , $-6$ for $b$ , $-3$ for $c$

  $x=-1+3t$

  $y=2-6t$

  $z=4-3t$

Thus, the parametric equations of the line through $A$ and $C$ are $x=-1+3t$ , $y=2-6t$ , and $z=4-3t$ .