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Math Help - vector problem!

  1. #1
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    vector problem!

    i apologise if this is the incorrect forum, i assume it would go under here or perhaps calculus forum - vector help!


    find the formula for ||b -a||^2 in terms of ||b||^2, ||a||^2 and ba

    thankyou so much in advance!
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  2. #2
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    Quote Originally Posted by iExcavate View Post
    i apologise if this is the incorrect forum, i assume it would go under here or perhaps calculus forum - vector help!


    find the formula for ||b -a||^2 in terms of ||b||^2, ||a||^2 and ba

    thankyou so much in advance!

    This question belongs, me believes ,in trigonometry-analytic geometry. Anyway:

    b-a is an arrow beginning at a and ending at b and \|b-a\| is its length, so using the cosines law (theorem) on the resulting triangle we get:

    \|b-a\|^2=\|a\|^2+\|b\|^2-2\|a\|\|b\|\cos \theta

    I'm not sure what is ba, but if you meant the dot product of these two vectors then just remember that \cos \theta=\frac{b\cdot a}{\|b\|\|a\|} , so you can substitute above.

    Tonio
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  3. #3
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    how do i deduce \|b-a\|^2 = \|b\|^2 + \|a\|^2 - 2\|a\|\|b\|cos\theta
    to prove: \cos \theta=\frac{b\cdot a}{\|b\|\|a\|}
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  4. #4
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    tonio told you: apply the cosine law to the triangle with sides a, b, and a- b.
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