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maw [93]
3 years ago
12

A particle starts at the point P=(−5,−1,3) when t=0 and moves along a straight line toward Q=(−6,−6,8) at a speed of 9 cm/sec. L

et x, y, and z be measured in cm, and t in seconds. Find a parametric vector equation for the position of the object.
Mathematics
1 answer:
navik [9.2K]3 years ago
6 0

Answer:

Step-by-step explanation:

Given

Particle moves from P\ (-5,-1,3)\ to\ Q\ (-6,-6,8)

speed of particle is v=9\ cm/s

Unit vector in the direction of vector \vec{PQ} is given by

\hat{n}=\frac{\vec{Q}-\vec{P}}{|\vec{Q}-\vec{P}|}

\hat{n}=\frac{-}{\sqrt{(-1)^2+(-5)^2+(5)^2}}

\hat{n}=\frac{}{\sqrt{1+25+25}}

Now Position of Particle at any time is given by

\vec{r(t)}=velocity\cdot time\cdot direction

\vec{r(t)}=9\cdot t\cdot \frac{}{\sqrt{1+25+25}}

Parametric vector equation is given by

x=\frac{-9t}{\sqrt{51}}

y=\frac{-45t}{\sqrt{51}}

z=\frac{45t}{\sqrt{51}}

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