How do you find the magnitude and direction angle of the vector $v=3(cos60^circi+sin60^circj)$ ?

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Answer 1 · 2017-02-08T09:30:04

$||vecv||=3; "the Direction Angle="60^@$ .

Observe that, $vec u=(cos60^@i+sin60^@j)$ is a Unit Vector,

making an angle of $60^@$ with the $+ve$ direction of X-Axis; &,

$vecv=3vecu.$

From this, we can immediately conclude that the magnitude of

$vecv=||vecv||=3$ , and, the Direction Angle $theta=60^@$ .

However, we may formally solve the Problem as follows :-

$vec v=3(cos60^@i+sin60^@j)=3{(1/2)i+(sqrt3/2)j}$

$=(3/2)i+((3sqrt3)/2)j$

$:. ||vecv||=sqrt{(3/2)^2+((3sqrt3)/2)^2}$

$=3/2sqrt(1+3)=3/2*2=3$ .

Suppose that the Direction Angle of $vecv$ is $theta.$

Then, $/_(vecv, veci)=theta.$

$rArr vecv.veci=||vecv||||veci||cos theta$

Here, $vecv=(3/2,(3sqrt3)/2), veci=(1,0)$

$(3/2)(1)+((3sqrt3)/2)(0)=(3)(1)costheta$

$rArr 3/2=3costheta rArr costheta=1/2$

$rArr theta=60^@$ .

How do you find the magnitude and direction angle of the vector v=3(cos60^circi+sin60^circj)v=3(cos60^circi+sin60^circj)?