Find a vector parametric equation \vec{r} (t) for the line through the points P = (-3,-2,-4) and...

Question:

Find a vector parametric equation {eq}\vec{r} (t){/eq} for the line through the points {eq}P = (-3,-2,-4){/eq} and {eq}Q = (-6, -1, -2){/eq} for each of the given conditions on the parameter {eq}t{/eq}.

(a) If {eq}\vec{r} (0) = \left \langle -3,-2,-4 \right \rangle{/eq} and {eq}\vec{r} (8) = \left \langle -6,-1,-2 \right \rangle{/eq}, then {eq}\vec{r} (t) = {/eq}

(b) If {eq}\vec{r} (4) = P{/eq} and {eq}\vec{r} (8) = Q{/eq}, then {eq}\vec{r} (t) = {/eq}

(c) If the points {eq}P{/eq} and {eq}Q{/eq} correspond to the parameter values {eq}t = 0{/eq} and {eq}t = -4{/eq}, respectively, then {eq}\vec{r} (t) = {/eq}

Parametric Curves of Line Segments


To find a parametrization of a line segment between two given points we will determine the direction vector of the line and

set up the starting and ending values of the parameter from the parametrization,

by taking multiples of the direction vector .

Answer and Explanation:

a) A parametric vector equation of the line passing through {eq}\displaystyle P=(-3, -2, -4) \text{ and } Q=(-6, -1, -2) , {/eq} such that {eq}\displaystyle \mathbf{r}(0)= \langle -3, -2, -4 \rangle, \mathbf{r}(8)=\langle -6, -1, -2\rangle, {/eq}

we will first determine the direction vector of the line, which is

{eq}\displaystyle \mathbf{u}=\vec{PQ}=\langle -6+3, -1+2, -2+4\rangle =\langle -3, 1, 2\rangle, {/eq}

The vector parametric equation is {eq}\displaystyle \mathbf{r}(t) =\langle -3-3at, -2-bt, -4+2ct\rangle, {/eq} where {eq}\displaystyle a, b,c {/eq} are determined from the condition of the end-point {eq}\displaystyle t=8 {/eq}

{eq}\displaystyle \begin{align}\mathbf{r}(8)&=\langle -6, -1, -2\rangle \iff \begin{cases} -3-24a=-6\\ -2-8b=-1\\ -4+16c=-2 \end{cases}\iff a=\frac{1}{8}, b=-\frac{1}{8}, c=\frac{1}{8}\\ &\text{ therefore, the equation is } \boxed{ \mathbf{r}(t)=\left\langle -3-\frac{3t}{8}, -2+\frac{t}{8}, -4+\frac{t}{4}\ \right\rangle, t\in\mathbf{R}}. \end{align} {/eq}

b) To find the parametric equation of the line passing through P and Q such that {eq}\displaystyle \mathbf{r}(4)= \langle -3,-2,-4 \rangle, \mathbf{r}(8)=\langle -6,-1,-2\rangle, {/eq}

we will use the vector parametric equation similar to a), but with t translated 4 units,

{eq}\displaystyle \mathbf{r}(t) =\langle -3-3a(t-4), -2-b(t-4), -4+2c(t-4)\rangle, {/eq} where {eq}\displaystyle a, b,c {/eq} are determined from the condition of the end-point {eq}\displaystyle t=8. {/eq}

{eq}\displaystyle \begin{align}\mathbf{r}(8)&=\langle -6,-1,-2\rangle \iff \begin{cases} -3-12a=-6\\ -2-4b=-1\\ -4+8c=-2 \end{cases}\iff a=\frac{1}{4}, b=-\frac{1}{4}, c=\frac{1}{4}\\ &\text{ therefore, the equation is } \mathbf{r}(t)=\left\langle -3-\frac{3(t-4)}{4}, -2+\frac{t-4}{4}, -4+\frac{t-4}{2}\right\rangle\iff \boxed{ \mathbf{r}(t)=\left\langle -\frac{3t}{4}, -3+\frac{t}{4},-6+\frac{ t}{2} \right\rangle, t\in\mathbf{R}}. \end{align} {/eq}

c) To find the parametric equation such that {eq}\displaystyle \mathbf{r}(0)= \langle -3,-2,-4 \rangle, \mathbf{r}(-4)=\langle -6,-1,-2\rangle, {/eq}

we will use the vector parametric equation given in a), one point corresponds to {eq}\displaystyle t=0 {/eq}

{eq}\displaystyle \mathbf{r}(t) =\langle -3-3at, -2-bt, -4+2ct \rangle, {/eq} where {eq}\displaystyle a, b, c {/eq} are determined from the condition of the point {eq}\displaystyle t=-4 {/eq}

{eq}\displaystyle \begin{align}\mathbf{r}(-4)&=\langle -6,-1,-2\rangle \iff \begin{cases} -3+12a=-6\\ -2+4b=-1\\ -4-8c=-2 \end{cases}\iff a=-\frac{1}{4}, b=\frac{1}{4}, c=-\frac{1}{4}\\ &\text{ therefore, the equation is } \boxed{ \mathbf{r}(t)=\left\langle -3+\frac{3t}{4}, -2-\frac{t}{4}, -4-\frac{t}{2} \right\rangle, t\in\mathbf{R}}. \end{align} {/eq}


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Graphs of Parametric Equations

from Precalculus: High School

Chapter 24 / Lesson 5
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