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Homework Help: Find a vector perpendicular to these vectors

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Find a vector perpendicular to these vectors

    a = i + 2j + 7k
    b = i + j - 2k

    2. Relevant equations
    a.b = |a||b|Cos(x)
    a x b = |a||b|Sin(x)

    3. The attempt at a solution

    So a.b = 0
    a x b = |a||b|

    |a| = sqrt(54)
    |b| = sqrt(6)

    a x b = sqrt(324)

    Have I acutally got anywhere?

    Thanks
    Tom
     
  2. jcsd
  3. Jan 6, 2010 #2

    Landau

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    This is nonsense. The LHS is a vector, while the RHS is a scalar. Correct is: |a x b|=|a||b|sin(x), where x is the angle between a and b.

    You are somewhat on the right track: the cross product a x b is a vector perpendicular to both a and b, so all you have to do is compute a x b. You have only calculated the norm of a x b, namely |a x b|=sqrt(324). But you want to compute the vector a x b itself!

    You should know how to do this. Try again.
     
  4. Jan 6, 2010 #3

    vela

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    [tex]a{\cdot}b[/tex] is not zero. Those two vectors aren't perpendicular to each other. You don't need to know that to solve this problem, but I thought it was worth mentioning.
     
  5. Jan 6, 2010 #4
    The cross product of two vector is perpendicular to the original vectors.

    [tex]\vec{a}X\vec{b}=\vec{c}\Rightarrow \vec{c}[/tex] perpendicular to [tex]\vec{a} [/tex] and [tex]\vec{b}[/tex]
     
  6. Jan 6, 2010 #5

    Landau

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    I am curious: why would you repeat something that has already been clearly mentioned?
     
  7. Jan 7, 2010 #6

    HallsofIvy

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    Here is your basic error. This is NOT "a x b", it is the length of a x b. a x b is a vector, not a number.

     
  8. Jan 11, 2010 #7
    Right how about using determinants. I get

    -11i + 9j - k

    and that is the right answer, but why is that vector perpendicular to both the other vectors? Why by multiplying vectors a and b together do I get a new vector with is perpendicular to both?

    Thanks
    Thomas
     
  9. Jan 11, 2010 #8
    does it have something to do with ijkijk (i x j = k)
     
  10. Jan 12, 2010 #9
    more specifically my question noew is

    why using the determinants
    i j k
    1 2 7
    1 1 -2

    Do I get a vector that is perpendicular to both of these vectors? Perpendicular where exactly? Is this a special case that a vector will be perpendicular to both vectors or is this true for any vector?

    Thanks
    Thomas
     
  11. Jan 12, 2010 #10
    It is a general rule that the cross product of two vectors (they can be whatever) gives a second vector prependicular to both the original vectors. So why prependicular? If we take a x b = c, then you can easily check a.(a x b) = a.c = 0 = b.(a x b)= b.c. Remember that if the cross product vanishes, then no such vector would be obtained.

    AB
     
  12. Jan 12, 2010 #11

    vela

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    I don't know of a reason other than to say it's a consequence of the definition of the cross product. Some definitions say that the cross product is the vector perpendicular to the other two, i.e. it's perpendicular by definition. Other definitions tell you how to compute the components of the cross product, and then you can prove that that resulting vector is perpendicular to the other two.
     
  13. Jan 12, 2010 #12
    To discuss the question "why is the cross product of a and b perpendicular to these initial two vectors" - can you not just take this intuitive step from the result? If you perform the cross product operation on any two vectors, which define a 2D plane, the result will always be perpendicular to this plane. It may help to try this out for simple cases, like define a and b to be directly along the x and Y axis respectively, and then prove to yourself that the cross product of these is along the Z direction. Hope this helps
     
  14. Jan 12, 2010 #13

    HallsofIvy

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    There are several different ways of defining the cross product. The simplest is: If [itex]\vec{u}= a_x\vec{i}+ a_y\vec{j}+ a_z\vec{k}[/itex] and [itex]\vec{v}= b_x\vec{i}+ b_y\ve{j}+ b_z\vec{k}[/itex] then [itex]\vec{u}\times\vec{v}= (a_yb_z- a_zb_y)\vec{i}- (a_xb_z- a_zb_x)\vec{j}+ (a_xb_y- a_yb_x)\vec{k}[/itex], which then leads to that (purely symbolic) "determinant" calculation.

    It should be easy to see that
    [tex]\vec{u}\cdot (\vec{u}\times\vec{v})= (a_xa_yb_z-a_xa_zb_y)- (a_ya_yb_z- a_ya_zb_x)[/tex][tex]+ a_za_xby_y- a_za_yb_x= (a_xa_yb_z- a_xa_yb_z)[/tex][tex]+ (-a_xa_zb_y+ a_xa_zb_y)+ (a_ya_zb_x- a_ya_zb_x)= 0[/tex]
    and that
    [tex]\vec{v}\cdot (\vec{u}\times\vec{v})= (b_xa_yb_z- b_xa_zb_y)- (b_ya_yb_z- b_ya_zb_x)[/tex][tex]+ b_za_xb_y- b_za_yb_x= 0[/tex]

    The fact that [itex]\vec{u}\times\vec{v}[/itex] is always perpendicular to both [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] is the main use of the cross product.
     
    Last edited by a moderator: Jan 12, 2010
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