sodni3

2021-05-29

Find the vector and parametric equations for the line segment connecting P to Q.
P(0, - 1, 1), Q(1/2, 1/3, 1/4)

Szeteib

Vector equation of a line segment joining the points with position vectors ${r}_{0}$ and ${r}_{1}$ is
$r=\left(1-t\right){r}_{0}+t{r}_{1}$
Where $t\in \left[0,1\right]$
Substitute ${r}_{0}=<0,-1,1>$ and ${r}_{1}=<\frac{1}{2}.\frac{1}{3},\frac{1}{4}>$
$r\left(t\right)=\left(1-t\right)\left(0,-1,1\right)+t<\frac{1}{2},\frac{1}{3},\frac{1}{4}>$
$r\left(t\right)=<0,-1+t,1-t>+<\frac{t}{2},\frac{t}{3},\frac{t}{4}>$
$r\left(t\right)=<\frac{t}{2},-1+\frac{4t}{3},1-\frac{3t}{4}>$
Where $t\in \left[0,1\right]$ The parametric equations for the line segment are
$x=\frac{t}{2},y=-1+\frac{4t}{3},z=1-\frac{3t}{4}$
Where $t\in \left[0,1\right]$

$x=\frac{t}{2}$
$y=-1+\frac{t}{3}$
$z=1+\frac{t}{4}$
Explanation:
To find the vector and parametric equations for the line segment connecting points P and Q, we can use the following steps:
1. Find the direction vector:
$\stackrel{\to }{v}=\stackrel{\to }{PQ}=\stackrel{\to }{Q}-\stackrel{\to }{P}$
2. Calculate the parametric equations:
$x={x}_{P}+t{v}_{x}$
$y={y}_{P}+t{v}_{y}$
$z={z}_{P}+t{v}_{z}$
Now, let's calculate the values using the given points P(0, -1, 1) and Q(1/2, 1/3, 1/4):
1. Direction vector:
$\stackrel{\to }{v}=\stackrel{\to }{Q}-\stackrel{\to }{P}=\left(\frac{1}{2},\frac{1}{3},\frac{1}{4}\right)-\left(0,-1,1\right)$
2. Parametric equations:
$x=0+t\left(\frac{1}{2}\right)$
$y=-1+t\left(\frac{1}{3}\right)$
$z=1+t\left(\frac{1}{4}\right)$

Eliza Beth13

Step 1. Calculate the vector $𝐯$ that represents the direction of the line segment. This can be done by subtracting the coordinates of point P from the coordinates of point Q:
$𝐯=𝐐-𝐏=\left(\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{4}\end{array}\right)-\left(\begin{array}{c}0\\ -1\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}+1\\ \frac{1}{4}-1\end{array}\right)=\left(\begin{array}{c}\frac{1}{2}\\ \frac{4}{3}\\ -\frac{3}{4}\end{array}\right).$
Step 2. Express the parametric equations for the line segment using the vector $𝐯$ and the coordinates of point P. Let's denote the parameter as $t$:
$𝐫\left(t\right)=𝐏+t𝐯=\left(\begin{array}{c}0\\ -1\\ 1\end{array}\right)+t\left(\begin{array}{c}\frac{1}{2}\\ \frac{4}{3}\\ -\frac{3}{4}\end{array}\right).$
The parametric equations can also be written component-wise as:
$x\left(t\right)=0+\frac{1}{2}t,\phantom{\rule{1em}{0ex}}y\left(t\right)=-1+\frac{4}{3}t,\phantom{\rule{1em}{0ex}}z\left(t\right)=1-\frac{3}{4}t.$
These equations describe the line segment connecting points P and Q.

Mr Solver

To find the vector and parametric equations for the line segment connecting points P and Q, we can use the following steps:
1. Find the vector $\stackrel{\to }{PQ}$, which represents the displacement from point P to point Q:
$\stackrel{\to }{PQ}=\left[\begin{array}{c}{Q}_{x}-{P}_{x}\\ {Q}_{y}-{P}_{y}\\ {Q}_{z}-{P}_{z}\end{array}\right]$
Substituting the coordinates of P and Q:
$\stackrel{\to }{PQ}=\left[\begin{array}{c}\frac{1}{2}-0\\ \frac{1}{3}-\left(-1\right)\\ \frac{1}{4}-1\end{array}\right]$
Simplifying:
$\stackrel{\to }{PQ}=\left[\begin{array}{c}\frac{1}{2}\\ \frac{4}{3}\\ -\frac{3}{4}\end{array}\right]$
2. The parametric equations for the line segment can be written as follows:
$x={P}_{x}+t·{\stackrel{\to }{PQ}}_{x}$
$y={P}_{y}+t·{\stackrel{\to }{PQ}}_{y}$
$z={P}_{z}+t·{\stackrel{\to }{PQ}}_{z}$
Substituting the coordinates of point P and the components of $\stackrel{\to }{PQ}$:
$x=0+t·\frac{1}{2}$
$y=-1+t·\frac{4}{3}$
$z=1+t·\left(-\frac{3}{4}\right)$
Simplifying, we get the parametric equations for the line segment:
$\overline{)x=\frac{t}{2}}$
$\overline{)y=-1+\frac{4t}{3}}$
$\overline{)z=1-\frac{3t}{4}}$
These equations describe the line segment connecting points P(0, -1, 1) and Q($\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$).

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