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Write the vector v in the form aiplusbj​, given its magnitude left norm Bold v right norm and the angle alpha it makes with the positive​ x-axis.Left norm Bold v right normequals12​, alphaequals150

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Answer:


6√(3)i-6j

Explanation:

Given:

The magnitude, v=12

The angle
\alpha it makes with the positive​ x-axis =
150^0

Now for a vector (x,y)


  • \alpha =tan^(-1)((y)/(x) ) ; and

  • v=√(x^2+y^2)

Therefore:


150 =tan^(-1)((y)/(x) )\\(y)/(x) =tan 150\\(y)/(x)=-(√(3) )/(3)


y=-(x √(3))/(3)

Similarly:


v=√(x^2+y^2)\\12=√(x^2+y^2)\\12^2=x^2+y^2\\144=x^2+y^2


144=x^2+(-(x √(3))/(3) )^2\\144=x^2+(3x^2)/(9) \\144=(4x^2)/(3)\\4x^2=144 X 3=432\\x^2=108\\x=6√(3)

Recall that:


y=-(x √(3))/(3)\\y=-(6√(3) √(3))/(3)=-6

Therefore:
(x,y)=(6√(3), -6)

The vector in the form ai+bj is therefore:


6√(3)i-6j

User Kaspa
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