How do you find the magnitude and direction angle of the vector $v=8(cos135^circi+sin135^circj)$ ?

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Answer 1 · 2018-08-03T13:18:26

Standard form:
magnitude ( cos ( direction angle ) $veci$ + sin ( direction angle ) $vecj )$ . Answer: 8, and $135^o in Q_2$ .

Graph for like-parallel $vec v$ , from O:
graph{(x^2+y^2-64)(y+0.707x)=0[-5.7 0 0 2.8]}

How do you find the magnitude and direction angle of the vector v=8(cos135^circi+sin135^circj)v=8(cos135^circi+sin135^circj)?