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    simplify (a+b).(a+b) given that |a| = 2 and |b| = 3

    the textbook says the answer is 13 + 2a.b, but can't I write that as 25? I know this is a really straightforward question but I'm just confused why 25 isn't their answer
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    How are you getting 25?

    a.b is not |ab|

    But a.a =|a|^2
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    You don't know the angle between a and b so you can't assume that a.b=12 because that would mean a and b are parallel, but you don't know if thats true.
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    (Original post by NotNotBatman)
    How are you getting 25?

    a.b is not |ab|

    But a.a =|a|^2
    understand now, thanks.
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    (Original post by NotNotBatman)
    How are you getting 25?

    a.b is not |ab|

    But a.a =|a|^2
    And what would |ab| equal?
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    (Original post by MR1999)
    And what would |ab| equal?
    6...
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    (Original post by MR1999)
    And what would |ab| equal?
    = |a||b| = 6
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    (Original post by NotNotBatman)
    = |a||b| = 6
    I've never come across that notation before. Interesting.
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    (Original post by MR1999)
    That would be |a||b|, unless he's using the notation that if u is a vector then u is the modulus of that vector. But in that case the absolute value signs around ab would be unnecessary as the value of ab would already be positive.
    a=\pm 2 and b = \pm 3

    So ab = \pm 6

    In any case, |ab| = 6 always.
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    (Original post by RDKGames)
    a=\pm 2 and b = \pm 3

    So ab = \pm 6

    In any case, |ab| = 6 always.
    a and b are vectors?
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    (Original post by MR1999)
    a and b are vectors?
    Yes in the question, but I'm just taking an example of scalars with those values in my post.

    As for vectors, you can show that |\mathbf{a}||\mathbf{b}| = |\mathbf{a}\cdot \mathbf{b}| quite easily if both are parallel.
    Though admittedly, it's too early for me and I missed out the fact that they are not necessarily parallel (whereas the dot product in question is indeed between two parallel vectors) so technically we can't say the above is true hence I take back the above; we would instead have |\mathbf{a}\cdot \mathbf{b}| = 6|\cos \theta| where \theta is the angle between \mathbf{a} and \mathbf{b}
 
 
 
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