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Math Help - addition: magnitude of a vector

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    addition: magnitude of a vector

    Given \mid \overline{a}\mid=3, \mid \overline{b}\mid=5 and \mid \overline{a}+\overline{b}\mid=7, determine \mid \overline{a}-\overline{b}\mid
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    Quote Originally Posted by juanfe_zodiac View Post
    Given \mid \overline{a}\mid=3, \mid \overline{b}\mid=5 and \mid \overline{a}+\overline{b}\mid=7, determine \mid \overline{a}-\overline{b}\mid
    Here's a hint, see where you can get with it

    | \overline a - \overline b |^2 = | \overline a |^2 - 2 \overline a \cdot \overline b + | \overline b |^2

    and

    | \overline a + \overline b |^2 = | \overline a |^2 + 2 \overline a \cdot \overline b + | \overline b |^2 \implies 2 \overline a \cdot \overline b = | \overline a + \overline b |^2 - | \overline a |^2 - | \overline b |^2
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    Quote Originally Posted by Jhevon View Post
    Here's a hint, see where you can get with it

    | \overline a - \overline b |^2 = | \overline a |^2 - 2 \overline a \cdot \overline b + | \overline b |^2

    and

    | \overline a + \overline b |^2 = | \overline a |^2 + 2 \overline a \cdot \overline b + | \overline b |^2 \implies 2 \overline a \cdot \overline b = | \overline a + \overline b |^2 - | \overline a |^2 - | \overline b |^2

    still lost...i dont know why you are raising them to the power of 2...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by juanfe_zodiac View Post
    still lost...i dont know why you are raising them to the power of 2...
    ...you were given numerical values for most of what you see in my post. raise them to 2 as you would any other number. for instance, wherever you see | \overline a | you can write 3 instead
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    Quote Originally Posted by Jhevon View Post
    ...you were given numerical values for most of what you see in my post. raise them to 2 as you would any other number. for instance, wherever you see | \overline a | you can write 3 instead

    is the answer 34??? i reemplaced 3 and 5 where the a's and the b's were thats what i got...i dont think im right...
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    Quote Originally Posted by juanfe_zodiac View Post
    is the answer 34??? i reemplaced 3 and 5 where the a's and the b's were thats what i got...i dont think im right...
    no, that's not right. what equation did you place them in?
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    Quote Originally Posted by Jhevon View Post
    no, that's not right. what equation did you place them in?
    \implies 2 \overline a \cdot \overline b = | \overline a + \overline b |^2 - | \overline a |^2 - | \overline b |^2

    i found a^2-b^2
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    Quote Originally Posted by juanfe_zodiac View Post
    \implies 2 \overline a \cdot \overline b = | \overline a + \overline b |^2 - | \overline a |^2 - | \overline b |^2
    first of all, you would not get 34 by plugging in the values we are given. secondly, you are looking for | \overline a - \overline b| not 2 \overline a \cdot \overline b
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    Quote Originally Posted by Jhevon View Post
    first of all, you would not get 34 by plugging in the values we are given. secondly, you are looking for | \overline a - \overline b| not 2 \overline a \cdot \overline b

    i dont know where am i going to get this | \overline a - \overline b| from those formulas...help me plzz
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    Quote Originally Posted by juanfe_zodiac View Post
    i dont know where am i going to get this | \overline a - \overline b| from those formulas...help me plzz
    i am helping. did you read my post? try to follow the logic here.

    i gave you the formula for |\overline a - \overline b|^2, you only need to square root both sides to find |\overline a - \overline b|.

    the problem is, the equation has 2 \overline a \cdot \overline b in it, which we know nothing about. my last equation shows how to calculate it using values we know so that we can plug it into the first equation
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    Quote Originally Posted by Jhevon View Post
    i am helping. did you read my post? try to follow the logic here.

    i gave you the formula for |\overline a - \overline b|^2, you only need to square root both sides to find |\overline a - \overline b|.

    the problem is, the equation has 2 \overline a \cdot \overline b in it, which we know nothing about. my last equation shows how to calculate it using values we know so that we can plug it into the first equation

    the answer was there all the time...thank you and sorry for wasting your time....i got the answer....algebra is just magic...(4.36)
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    Quote Originally Posted by juanfe_zodiac View Post
    the answer was there all the time...thank you and sorry for wasting your time....i got the answer....algebra is just magic...(4.36)
    leave it as \sqrt {19}
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    is up to his old tricks again! Jhevon's Avatar
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    i gave this problem to someone today and i came up with a new solution. this one is geometrical.

    i won't say what the solution is, i will just say it involves using the law of cosines with the diagram below as a guide.
    Attached Thumbnails Attached Thumbnails addition: magnitude of a vector-vector-diagram.bmp  
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    Quote Originally Posted by Jhevon View Post
    i gave this problem to someone today and i came up with a new solution. this one is geometrical.

    i won't say what the solution is, i will just say it involves using the law of cosines with the diagram below as a guide.

    someone also used a metod that had to deal with coordenates...xyz...but basicly is the same thing as what you did...but he used coordinates because he said that we cannot do the factorization thing because we are trying to find magnitude, a number and we cant spilt it the vectors like that because is a number...but he got the same answer as you did...thats weird...thanks for the new method
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    Quote Originally Posted by juanfe_zodiac View Post
    someone also used a metod that had to deal with coordenates...xyz...but basicly is the same thing as what you did...but he used coordinates because he said that we cannot do the factorization thing because we are trying to find magnitude, a number and we cant spilt it the vectors like that because is a number...but he got the same answer as you did...thats weird...thanks for the new method
    i'd like to see that solution. coming up with coordinates for this seems like a pain. it is much easier to just draw an informal diagram as i did. you get to the same conclusion with less work. to come up with the coordinates, you probably have to do some of the things i did here anyway.
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