Khaleesi Herbert

2021-09-12

Represent the line segment from P(−2, −3, 8), Q(5, 1, −2) by a vector-valued function and by a set of parametric equations.

Layton

We have, points P(−2, −3, 8) and Q(5, 1, −2)
Now, the component vector from P to Q ,
$v\left(t\right)=<5-\left(-2\right),1-\left(-3\right),-2-8>⇒v\left(t\right)=<7,4,-10>$
So, the direction number are a=7,b=4 and c=−10
For initial point on the line, use point $P\left(-2,-3,8\right)$
We get,${x}_{0}=-2,y0=-3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{z}_{0}=8$
Now, to form the parametric equation:
$x={x}_{0}+at,y={y}_{0}+bt\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}z={z}_{0}+ct\dots \dots .\left(1\right)$
Putting the value of $a=7,b=4,c=-10$ and ${x}_{0}=-2,{y}_{0}=-3{z}_{0}=8$ in eq.(1),we get $x=-2+7t,y=-3+4t$ and $z=8-10t$
Now, to form a vector−valued function: $r\left(t\right)=x\left(t\right)i+y\left(t\right)j+z\left(t\right)k$
We have, $x=-2+7t,y=-3+4t$ and $z=8-10t$

So, $r\left(t\right)=\left(7t-2\right)i+\left(4t-3\right)j+\left(8-10t\right)k$
Hence, set of parametric equations be ,$x=-2+7t,y=-3+4t$ and $z=8-10t$ and vector−valued function be, $r\left(t\right)=\left(7t-2\right)i+\left(4t-3\right)j+\left(8-10t\right)k$

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