Wribreeminsl
2021-02-02
Answered

Find sets of parametric equations and symmetric equations of the line that passes through the given point and is parallel to the given vector or line. (For each line, write the direction numbers as integers.)
Point $(-1,\text{}0,\text{}8)$
Parallel to
$v=3i\text{}+\text{}4j\text{}-\text{}8k$
The given point is $(-1,\text{}0,\text{}8)\text{}\text{and the vector or line is}\text{}v=3i\text{}+\text{}4j\text{}-\text{}8k.$
(a) parametric equations
(b) symmetric equations

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jlo2niT

Answered 2021-02-03
Author has **96** answers

(a)
The parametric equations for a line passing through $({x}_{0},\text{}{y}_{0},\text{}{z}_{0})$ and parallel to the vector
$v=ai\text{}+\text{}bj\text{}+\text{}ck$ are
$x={x}_{0}\text{}+\text{}at,\text{}y={y}_{0}\text{}+\text{}bt,\text{}z={z}_{0}\text{}+\text{}ct$
The required parametric equations are
$x=\text{}-1\text{}+\text{}3t,\text{}y=4t,\text{}z=8\text{}-\text{}8t$
(b)
The parametric equations are
$x=\text{}-1\text{}+\text{}3t,\text{}y=4t,\text{}z=8\text{}-\text{}8t$
Solving for t we have,
$\Rightarrow \text{}t=\text{}\frac{x\text{}+\text{}1}{3},\text{}t=\text{}\frac{y}{4},\text{}t=\text{}\frac{z\text{}-\text{}8}{-8}$

$\Rightarrow \text{}\frac{x\text{}+\text{}1}{3}=\text{}\frac{y}{4}=\text{}\frac{z\text{}-\text{}8}{-8}$

$\Rightarrow \text{}\frac{x\text{}+\text{}1}{3}=\text{}\frac{y}{4}=\text{}\frac{8\text{}-\text{}z}{8}$
Symmetric equations
$\frac{x\text{}+\text{}1}{3}=\frac{y}{4}=\frac{8\text{}-\text{}z}{8}$

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