2 people move in the same direction along a line of equation:

(x + 3)/10 = (y + 10)/20 = (z - 10)/-20

Mobile M1 is at a point A(-3, -10, 10) and moves at a velocity of 6m/s towards point B(7, 10, -10) where Mobile 2 is found, walking in the same direction at a velocity of 3m/s. Suppose that the line equation and the coordinates are expressed in meters:

a) For each mobile, give a vector equation to determine the position at each instant t, where t is in seconds.

3 answers

To find the vector equation of M1, we need to find the direction along which it is moving, which is given by the difference of coordinates of B and A:
BA=<7-(-3),10-(-10),-10-10>
=<10,20,-20>
We know that the speed (magnitude) is 3 m/s, so we normalize the direction vector and multiply by 3.
The unit vector is obtained by dividing the vector by its magnitude:
|BA|=sqrt(10^2+20^2+20^2)=30
Therefore unit vector of BA,
ba=<10/30,20/30,-20/30>=<1/3, 2/3, -2/3>
For a speed of 3 m/s, we multiply the unit vector by 3:
M1=3ba=<1,2,-2>
Since M1 start from (-3,-10,10) at time t=0, we have the vector equation as
x=<-3+t, -10+2t, 10-2t>

M2 can be calculated similarly.

THanks Mate!

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