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Math Help - Proving Vectors

  1. #1
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    Proving Vectors

    A) By considering the angles between the vectors, show that a+b and
    a-b are perpendicular when |a|=|b|

    Note: all bolded above are vectors (arrows on top)

    Below is my attempt on the above questions, please let me know if I have the correct answer (as there are no answers in the back for these questions). Demonstrate any errors that I may have, the correct way. Thanks in advance!

    A) well I drew two right angled triangles
    the first has a as the base, b as the height and hypotenuse is a+b, the second has a as base, -b as height, and hypotenuse is a-b
    .. not sure about this
    Last edited by skeske1234; August 24th 2009 at 07:42 AM.
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    A) By considering the angles between the vectors, show that a+b and
    a-b are perpendicular when |a|=|b|
    \left( {\overrightarrow a  + \overrightarrow b } \right) \cdot \left( {\overrightarrow a  - \overrightarrow b } \right) = \overrightarrow a \overrightarrow { \cdot a}  - \overrightarrow b  \cdot \overrightarrow b

    \overrightarrow a \overrightarrow { \cdot a}  = \left\| {\overrightarrow a } \right\|^2
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    what happens to the bb?
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    Quote Originally Posted by skeske1234 View Post
    what happens to the bb?
    He thought you could work that out for yourself! If, as Plato says, \vec{a}\cdot\vec{a}= |\vec{a}|^2,l what do you think \vec{b}\cdot\vec{b}?

    Do you see what happened to \vec{a}\cdot\vec{b} and \vec{b}\cdot\vec{a}?

    And, of course, a condition for this problem is that |\vec{a}|= |\vec{b}|.

    The problem with your attempted proof of (A) is that you started with a right triangle: assuming that \vec{a}+ \vec{b} and \vec{a}- \vec{b} are perpendicular. You are asked to prove the opposite: that if |\vec{a}|= |\vec{b}| then \vec{a}- \vec{b} and \vec{a}+ \vec{b} are perependicular. You were "going the wrong way".
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    So is this right then:
    (a+b) (a-b)= aa - bb
    =|a|^2 - |b|^2

    and how does this answer prove that a+b and a-b are perpendicular?
    if |a|=|b| then wouldnt the answer be 0? hm.. so zero is equal to |a+b| and |a-b| .. how does that determine that they are perpendicular though?
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    Quote Originally Posted by skeske1234 View Post
    So is this right then:
    (a+b) (a-b)= aa - bb
    =|a|^2 - |b|^2

    and how does this answer prove that a+b and a-b are perpendicular?
    if |a|=|b| then wouldnt the answer be 0? hm.. so zero is equal to |a+b| and |a-b| .. how does that determine that they are perpendicular though?
    If the dot product of two vectors is 0, then the two vectors are perpendicular.
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  7. #7
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    Quote Originally Posted by Plato View Post
    \left( {\overrightarrow a  + \overrightarrow b } \right) \cdot \left( {\overrightarrow a  - \overrightarrow b } \right) = \overrightarrow a \overrightarrow { \cdot a}  - \overrightarrow b  \cdot \overrightarrow b

    \overrightarrow a \overrightarrow { \cdot a}  = \left\| {\overrightarrow a } \right\|^2
    I don't know what the question means by "considering the angles between the vectors", though...

    The dot product of two vectors x and y is:

    |x| |y| cos(theta)

    Where theta is the angle between them. When the dot product is zero, and |x| and |y| are different from 0, that means cos(theta) = 0, which for 0<theta<pi means that theta=pi/2 --> perpendicular.
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    Quote Originally Posted by pedrosorio View Post
    I don't know what the question means by "considering the angles between the vectors", though...

    The dot product of two vectors x and y is:

    |x| |y| cos(theta)

    Where theta is the angle between them. When the dot product is zero, and |x| and |y| are different from 0, that means cos(theta) = 0, which for 0<theta<pi means that theta=pi/2 --> perpendicular.
    Remember that

    \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}.

    If \mathbf{a} and \mathbf{b} are perpendicular, then \theta = 90^{\circ}, and so \cos{\theta} = 0.


    Thus \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|(0) = 0.


    So if the dot product of two vectors is 0, then they are perpendicular.
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