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Homework Help: Really fundamental cross/dot product questions

  1. Sep 6, 2014 #1
    So I've been working on physics homework and we have some vector/dot product questions.
    This is really long, but the questions I have really are rudimentary at best.

    I have seven total questions.

    You're given two vectors that only have an x and y component, A, and B, and the positive Z axis is out of the page.

    A points roughly (no numbers given) 45 degrees (north-east). B point roughly 315 degrees (north-west).

    I've tried to read the class notes on vectors and things, however I really can't seem to understand it.

    There are basically three things I'm trying to understand.
    Vectors, which I understand, and the difference between cross products and dot products.

    Cross products seem to be |a||b|sine([itex]\theta[/itex]).
    Dot products seem to be abCosine([itex]\theta[/itex]).

    So please explain if what I'm doing is right-

    What direction is vector axb? (A times B)
    So since it's a cross product, I guess it's the absolute magnitude of a and b times sine.
    However I really don't understand the angle thing. How is the angle measured? Is the [itex]\theta[/itex] measured depending on the x or y axis? Or just THE angle between A and B?
    So I wrote down the direction is upwards with no x-component? I have no better guess. Approximately the angle between a and b looks to be 90-120 degrees, 90 degrees, which would appear to sort of "add" a and b, just with larger magnitude.

    The next question is how is bxa related to axb in magnitude and direction?
    I'm sure that I'm wrong, however i can only think that cross products are commutative and that they are both identical. I guess there could be a thing with the angle, but still I'm not sure.

    How could you change the direction of a, leaving it in the x, y plane to make axb zero?
    Sine(180)=0, so you would just make a paralell to B.

    If we rotate the coordinate axis so that x points towards the bottom of the paper, which I assume means to rotate the grid by 90 degrees clockwise, "how will axb change?"
    This really doesn't seem to be relevant? I'm supposing a and b don't move with the graph, but either ways since first of all cross products are absolute values and that the angle between a and b is the same no matter how you rotate the xy axis, 27 seems to be a pointless question.

    I think I sort of understand dot products after reading the class notes.
    So one question I have is the result of a dot product, such as a[itex]\bullet[/itex]b, is C, which is a scalar. So a scalar is just a number, but I'm confused what the difference between a scalar and a vector is. A scalar has an angle and a magnitude, I think, so isn't that just a vector?

    And the only way to get a negative dot product is with the angle>90, right? Since a and b are both magnitudes, which cannot be negative?

    28. How could you change the direction of a, leaving it in the xy plane to make a dot b as large as possible?
    Since cosine 0 or 180= maximum, making A to be parallel would give the largest dot product.

    How could you change a's direction to make a dot b as small as possible? Can c be zero when a and b are not?
    When a is perpendicular to b, axb is zero. C can be zero any time a and b are nonzero but perpendicular.

    30 Can you assign a direction to C? If so, what direction would c be for the vectors a and b as shown?
    Since A is roughly 45 degrees and B is 315 degrees roughly, the dot product is close to zero. I am not sure if C is just a number or has an angle. I think it doesn't?
  2. jcsd
  3. Sep 6, 2014 #2
    You need to read up on the differences between scalars and vectors. You also need to read about cross products.

    A scalar is just a (real) number.

    A vector has both magnitude and direction. You should think of it as an arrow in the coordinate system. It has a certain length (its magnitude) and it points somewhere (its direction).

    The dot product takes two vectors and produces a scalar.

    The cross product takes two vectors and produces a new vector.

    The formula you have for the cross product is only for its magnitude. ##|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin \theta##, where ##\theta## indeed is the angle between the vectors ##\mathbf{a}## and ##\mathbf{b}##.

    Your book should also tell you something about how you determine the direction of the vector produced by the cross product.
  4. Sep 6, 2014 #3


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    Since this is the homework forum, I can only give you hints, not complete answers.

    The only part of that that's relevant here is that they're not parallel, and not 0.

    The dot product ##x\cdot y## is ##|x||y|\cos\theta## where ##\theta## is the angle between the vectors, but the cross product ##x\times y## is a vector with magnitude ##|x||y|\sin\theta##.

    Note that the definitions make it clear that ##x\cdot y## is a number and ##x\times y## is a vector:
    &x\cdot y=x_1y_1+x_2y_2+x_3y_3\\
    &x\times y=(x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_3)

    That angle only contributes to the magnitude, not the direction. The direction is always perpendicular to the plane that contains x and y. A lot of people use a "right-hand rule" to remember which of the two perpendicular directions it is. http://en.wikipedia.org/wiki/Right-hand_rule

    The angle has nothing to do with coordinate axes. There's a unique plane that contains x,y and 0. The angle is measured in that plane.

    You have a formula for the magnitude, so I suggest that you use it. Cross products aren't commutative. They're not even associative. Use the definition that I included above, or the right-hand rule, to figure out the direction.

    That's a good start, but there are actually two directions that work (two angles).

    The point is that the axes are irrelevant, or equivalently, that your choice of basis for ##\mathbb R^3## is irrelevant. This isn't obvious, since the definition I included above can give the impression that the basis matters. The numbers ##x_1,x_2,x_3,y_1,y_2,y_3## all depend on it.

    Scalars don't have angles (except when we're talking about complex scalars, but there's no need to talk about that now). A scalar is a number. A vector in ##\mathbb R^3## is an ordered triple of numbers ##(x_1,x_2,x_3)##.
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