Respuesta :
Answer:
Step-by-step explanation:
Part 1)
Since speed defined as the magnitude of velocity vector we have
[tex]|v|=\sqrt{v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}[/tex]
Applying values we get
[tex]|v|=\sqrt{2^{2}+3^{2}+5^{2}}[/tex]
[tex]\therefore v=6.16m/s[/tex]
Thus the speed of the object is 6.16 m/s
Part 2)
A unit vector in the direction of [tex]\overrightarrow{v}=v_{x}\widehat{i}+v_{y}\widehat{j}+v_{z}\widehat{k}[/tex] is given by
[tex]\frac{v_{x}\widehat{i}+v_{y}\widehat{j}+v_{z}\widehat{k}}{|v|}[/tex]
Applying values we get
Unit vector in direction of the given vector is
[tex]\frac{2\widehat{i}+3\widehat{j}+5\widehat{k}}{6.16}\\\\0.324\widehat{i}+0.487\widehat{j}+0.811\widehat{k}[/tex]
The unit vector in opposite direction is
[tex]-0.324\widehat{i}-0.487\widehat{j}-0.811\widehat{k}[/tex]
Part 3)
Since the speed of the particle reduces to 3 m/s from 6.16 m/s this can be brought about by varying each [tex]v_{x},v_{y},v_{z}[/tex] thus the problem is indeterminate since the no. of variables is 3 ( each individual component of velocity can be changed independently) but the equation avaliable is only 1. Thus we cannot find the velocity vector only by the information given.
