Question

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. Let x, y, and z be measured in cm, and t in seconds. Find a parametric vector equation for the position of the object.

Answer 1

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

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

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

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