Chapter 8.6, Problem 46E

Solution

To find: The parametric equations for the line through $C$ and the midpoint of $AB$ .

The parametric equations for the line through $C$ and the midpoint of $AB$ are $x=2-\frac{5}{2}t$ , $y=-4+8t$ , and $z=1-\frac{1}{2}t$ .

Given information:

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

Formula used:

Midpoint formula: The midpoint $M$ of the line segment $PQ$ with endpoints $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ is $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$ .

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

Where $t$ is a real number.

Calculation:

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

Let $M$ be the midpoint of $AB$ .

Using the formula $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substitute $-1$ for $x_1$ , $2$ for $y_1$ , $4$ for $z_1$ , $0$ for $x_2$ , $6$ for $y_2$ and $-3$ for $z_2$ .

  $\begin{array}{l}M=\left(\frac{-1+0}{2},\frac{2+6}{2},\frac{4-3}{2}\right)\\ M=\left(\frac{-1}{2},\frac{8}{2},\frac{1}{2}\right)\\ M=\left(\frac{-1}{2},4,\frac{1}{2}\right)\end{array}$

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

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

Substitute $2$ for $a_1$ , $-4$ for $a_2$ , $1$ for $a_3$ , $\frac{-1}{2}$ for $b_1$ , $4$ for $b_2$ and $\frac{1}{2}$ for $b_3$ .

  $\begin{array}{c}\overrightarrow{CM}=\left(\frac{-1}{2}-2,4-\left(-4\right),\frac{1}{2}-1\right)\\ \text{v}=\left(\frac{-5}{2},8,\frac{-1}{2}\right)\end{array}$

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

Substitute $2$ for $x_0$ , $-4$ for $y_0$ , $1$ for $z_0$ , $\frac{-5}{2}$ for $a$ , $8$ for $b$ , $\frac{-1}{2}$ for $c$

  $x=2-\frac{5}{2}t$

  $y=-4+8t$

  $z=1-\frac{1}{2}t$

Thus, the parametric equations of the line through $C$ and the midpoint of $AB$ are $x=2-\frac{5}{2}t$ , $y=-4+8t$ , and $z=1-\frac{1}{2}t$ .