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    When you work out the dot/ cross product, what are you actually working out? For the dot product, you take two vectors and end up with a scalar number...what does this number represent?
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    (Original post by Megan_101)
    When you work out the dot/ cross product, what are you actually working out? For the dot product, you take two vectors and end up with a scalar number...what does this number represent?
    The dot product of \mathbf{a} and \mathbf{b} is the length of the projection of \mathbf{a} onto \mathbf{b} multiplied by the length of \mathbf{b} (or the other way around--it's commutative).

    Pretty picture:

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    (Original post by Megan_101)
    When you work out the dot/ cross product, what are you actually working out? For the dot product, you take two vectors and end up with a scalar number...what does this number represent?
    The dot product :
     \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta .
     \cos \theta is the cosine of the angle between the 2 vectors.

    The cross product :
     \displaystyle \mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}|\mathbf{\hat{n}} \sin \theta .

     \theta is the angle between the vectors a and b and  \mathbf{\hat{n}} is the unit normal vector - a vector that has magnitude 1 and is perpendicular to both vectors a and b.
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    (Original post by Megan_101)
    When you work out the dot/ cross product, what are you actually working out? For the dot product, you take two vectors and end up with a scalar number...what does this number represent?
    On a slightly different note, if you're working with physical quantities, it's nicer to think of dot products and cross products in forms of tensors. You might want to look it up, if you're vaguely interested. The gist of it is that there are two tensors which are invariant under rotations: \delta_{ij} and \epsilon_{ijk} and you use these to define the dot product and cross product.
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    (Original post by Zacken)
    The dot product of \matbf{a} and \mathbf{b} is the length of the projection of \mathbf{a} onto \mathbf{b} multiplied by the length of \mathbf{b} (or the other way around--it's commutative).

    Pretty picture:

    excellent!
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    (Original post by B_9710)
     \mathbf{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta .
    Not supposed to have arrows but I don't know how to do bold vectors - so there you have it.
    Θ is the angle between the 2 vectors.
    What's wrong with arrows?

    It's \mathbf{a} and \mathbf{b}.
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    In physics, the work done by a force is defined as the dot product between the force acting on an object and the displacement the object does.
    Using cross product we can define torque and angular momentum

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    (Original post by depymak)
    In physics, the work done by a force is defined as the dot product between the force acting on an object and the displacement the object does.
    Which is, by the way, to anybody else reading why work done is a scalar even though force and displacement is a vector quantity. Taking their dot product results in a scalar, hence why work done is a scalar as well.
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    (Original post by Megan_101)
    When you work out the dot/ cross product, what are you actually working out? For the dot product, you take two vectors and end up with a scalar number...what does this number represent?
    The dot product is the "in-common-ness" of two vectors.

    The cross product of two vectors is a vector perpendicular to the plane formed by the linear span of the two vectors.
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    (Original post by Don John)
    The cross product of two vectors is a vector perpendicular to the plane formed by the linear span of the two vectors.
    I'm curious. Does this assume that the vector is three dimensional?
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    (Original post by Don John)
    The dot product is the "in-common-ness" of two vectors.

    The cross product of two vectors is a vector perpendicular to the plane formed by the linear span of the two vectors.
    Like it.
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    (Original post by Zacken)
    I'm curious. Does this assume that the vector is three dimensional?
    If you look at the matrix determinant method of calculating the cross-product, you'll see that one needs two vectors defined in 3D space. They don't have to actually have >0 values in all three dimensions.
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    (Original post by Don John)
    If you look at the matrix determinant method of calculating the cross-product, you'll see that one needs two vectors defined in 3D space. They don't have to actually have >0 values in all three dimensions.
    Not sure what you mean? You can have a cross-product in seven dimensions as well.
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    (Original post by Zacken)
    Not sure what you mean? You can have a cross-product in seven dimensions as well.
    I was talking about the usual case of finding the cross-product of two vectors to help the OP visualise the cross-product in their heads.

    Attempting to visualise the seven-dimension cross-product, while indeed a mathematically well-known concept, is a facetious exercise.
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    (Original post by Don John)
    I was talking about the usual case of finding the cross-product of two vectors to help the OP visualise the cross-product in their heads.

    Attempting to visualise the seven-dimension cross-product, while indeed a mathematically well-known concept, is a facetious exercise.
    Do you mean that we can't talk about the cross product if we are considering a two-dimensional space?
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    (Original post by depymak)
    Do you mean that we can't talk about the cross product if we are considering a two-dimensional space?
    Indeed.
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    (Original post by depymak)
    Do you mean that we can't talk about the cross product if we are considering a two-dimensional space?
    The two-dimensional equivalent of a cross product is a scalar:

    \mathbf{x} \times \mathbf{y} = x_1 y_2  - x_2 y_1

    Unlike dot products, cross products aren't geometrically generalisable to n dimensions.
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    (Original post by Zacken)
    The dot product of \mathbf{a} and \mathbf{b} is the length of the projection of \mathbf{a} onto \mathbf{b} multiplied by the length of \mathbf{b} (or the other way around--it's commutative).

    Pretty picture:

    Okay yep I understand that so the product of the magnitude of the component of A being mapped onto B (|a| cos theta) and the magnitude of B shows us what exactly geometrically?
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    (Original post by B_9710)
     \mathbf{\hat{n}} is the unit normal vector - a vector that has magnitude 1 and is perpendicular to both vectors a and b.
    Does  \mathbf{\hat{n}} always have a magnitude of 1?
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    (Original post by Megan_101)
    Does  \mathbf{\hat{n}} always have a magnitude of 1?
    Yes that's what the hat above the n means.
    When you simply find the cross product of 2 non parallel vectors, say a and b your answer gives a vector that is perpendicular to both a and b, but this perpendicular vector does not necessarily have a magnitude of 1.
 
 
 
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