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anonymous

  • one year ago

Find the magnitude & direction of the vector JK given points J(1,0) & K(-2,3).

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  1. jim_thompson5910
    • one year ago
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    First we need to find the vector in <x,y> form Subtract the corresponding coordinates (tail - head) x = -2 - 1 = -3 y = 3 - 0 = 3 the vector JK is <-3, 3> now you will use the formula sqrt(x^2 + y^2) to get the magnitude to get the direction, use arctan(y/x). The angle is in Q2, so you'll have to add 180 degrees

  2. jdoe0001
    • one year ago
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    \(\large u=<a,b>\qquad \begin{cases} magnitude\implies \sqrt{a^2+b^2}\\ direction\implies tan^{-1}\left( \frac{b}{a} \right) \end{cases}\)

  3. anonymous
    • one year ago
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    wait im kinda confused

  4. jim_thompson5910
    • one year ago
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    where are you stuck?

  5. anonymous
    • one year ago
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    nevermind i got it thanks

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