Question 1087028
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The magnitude of the vector 2v is {{{sqrt(4^2 + (-10)^2)}}} = {{{sqrt(16 + 100)}}} = {{{sqrt(116)}}} = {{{2*sqrt(29)}}}.


The direction angle of the vector 2v is the same as that of the vector v:

{{{tan(alpha)}}} = {{{(-5)/2}}} = -2.5.


Notice that the angle {{{alpha}}} lies in QIV.



Therefore,  {{{alpha}}} = {{{arctan(-2.5)}}},  or, which is the same,  {{{alpha}}} = {{{-arctan(2.5)}}}.
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