Respuesta :
Answer:
Step-by-step explanation:
Given
Particle moves from [tex]P\ (-5,-1,3)\ to\ Q\ (-6,-6,8)[/tex]
speed of particle is [tex]v=9\ cm/s[/tex]
Unit vector in the direction of vector [tex]\vec{PQ}[/tex] is given by
[tex]\hat{n}=\frac{\vec{Q}-\vec{P}}{|\vec{Q}-\vec{P}|}[/tex]
[tex]\hat{n}=\frac{<-5,-1,3>-<-6,-6,8>}{\sqrt{(-1)^2+(-5)^2+(5)^2}}[/tex]
[tex]\hat{n}=\frac{<-1,-5,5>}{\sqrt{1+25+25}}[/tex]
Now Position of Particle at any time is given by
[tex]\vec{r(t)}=velocity\cdot time\cdot direction[/tex]
[tex]\vec{r(t)}=9\cdot t\cdot \frac{<-1,-5,5>}{\sqrt{1+25+25}}[/tex]
Parametric vector equation is given by
[tex]x=\frac{-9t}{\sqrt{51}}[/tex]
[tex]y=\frac{-45t}{\sqrt{51}}[/tex]
[tex]z=\frac{45t}{\sqrt{51}}[/tex]
The parametric vector equation for the position of the object[tex]\vec{r} = \frac{-9ti}{\sqrt{51} } -\frac{5tj}{\sqrt{51} } \frac{+5tk}{\sqrt{51} }[/tex]
If a particle starts at the point P=(−5,−1,3) when t=0 and moves along a straight line toward Q=(−6,−6,8), the unit vector in the direction of PQ is expressed as:
[tex]e=\frac{PQ}{|PQ|}[/tex]
PQ is the distance from P to Q and is expressed as Q - P
PQ = Q - P = (−6,−6,8) - (−5,−1,3)
PQ = (-1, -5, 5)
Get the modulus of PQ;
|PQ| = √(-1)² + (-5)² + 5²
|PQ| = √1 + 25 + 25
|PQ| = √51
The unit vector in the direction of PQ is expressed as [tex]\uvec{e} =\frac{-i-5j+5k}{\sqrt{51} }[/tex]
Get the position vector [tex]\bm{\vec{r}}[/tex]
[tex]\bm{\vec{r}} = velocity \cdot time \cdot e[/tex]
velocity = 9m/s
Substitute the given parameters into the position vector as shown:
[tex]\vec{r}= 9t \cdot\frac{-i-5j+5k}{\sqrt{51} }\\\vec{r} = \frac{-9ti}{\sqrt{51} } -\frac{5tj}{\sqrt{51} } \frac{+5tk}{\sqrt{51} }[/tex]
Hence the parametric vector equation for the position of the object is[tex]\vec{r} = \frac{-9ti}{\sqrt{51} } -\frac{5tj}{\sqrt{51} } \frac{+5tk}{\sqrt{51} }[/tex]
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