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What is the dot product? watch

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    When you work out the dot product, what are you actually working out? Where does it come from? And why do we use it? I'm fine with all the calculations in the C4 book but it's just understanding what it is that I have trouble with.

    Thanks.
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    nyxnko_ been a while since I've done this so I can't really give a perfect answer. Best I can give is that its the multiplication of the projection of a vector a on a vector b (how much a acts in the direction of b) with the vector b (talking about the magnitudes), e.g. if b was a force applied in the horizontal, the dot product in the direction of b would be b multiplied by how much the force a acts in the horizontal. The equation is a*b*cos(theta) where a*cos(theta) is the projection of a on vector b (how much a acts in the direction of b, get this from right angled triangles) and then b is just the magnitude of b. One use of it would be to find the angle between two vectors where you have the displacements of the vectors to find the dot product and you can then rearrange with the magnitudes as in the dot product equation.

    https://www.youtube.com/watch?v=KDHu...ature=youtu.be
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    (Original post by Vikingninja)
    nyxnko_ been a while since I've done this so I can't really give a perfect answer. Best I can give is that its the multiplication of the projection of a vector a on a vector b (how much a acts in the direction of b) with the vector b (talking about the magnitudes), e.g. if b was a force applied in the horizontal, the dot product in the direction of b would be b multiplied by how much the force a acts in the horizontal. The equation is a*b*cos(theta) where a*cos(theta) is the projection of a on vector b (how much a acts in the direction of b, get this from right angled triangles) and then b is just the magnitude of b. One use of it would be to find the angle between two vectors where you have the displacements of the vectors to find the dot product and you can then rearrange with the magnitudes as in the dot product equation.

    https://www.youtube.com/watch?v=KDHu...ature=youtu.be
    Thanks for replying
    I think I understand what the dot product is now but I'm still a bit iffy about what it actually represents... like what the scalar product actually represents and why it's used...
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    (Original post by nyxnko_)
    Thanks for replying
    I think I understand what the dot product is now but I'm still a bit iffy about what it actually represents... like what the scalar product actually represents and why it's used...
    Why it's used in C4 should be clear - you can use it to find the angle between two vectors and also show that two vectors are perpendicular easily. It is defined as the product of the magnitudes divided by the cosine of the angle between the vectors.

    By "represent", I think you want to be able to visualise the dot product, just like you can visualise what the product of two numbers means. People try to give visualisations of the dot product but students don't usually like them because they're not immediately obvious - a product of two things should be simple they would say. But that's only because they're used to simple things like addition and multiplication of numbers. Why can't the dot product just be "product of the magnitudes divided by the cosine of the angle between the vectors"?

    Here's some geometric visualisations but you may not like them:

    If \mathbf{a} and \mathbf{b} are two vectors and \mathbf{b} is of unit length then \mathbf{a}\cdot \mathbf{b} is the projection of the vector \mathbf{a} on to \mathbf{b}. If you know mechanics, this means the component of \mathbf{a} in the direction of \mathbf{b}:





    But this only works if \mathbf{b} is a unit vector so you may think this is a bit crap. If \mathbf{b} is not a unit vector then \mathbf{a}\cdot \mathbf{b} is the projection of \mathbf{a} on to \mathbf{b} multiplied by the magnitude of \mathbf{b}. Nice? Not really.

    If you know about "work" in mechanics, this is the product of force (in direction of travel) and distance. If you have a force vector \mathbf{F} and a displacement vector \mathbf{r} then the work done by that force is \mathbf{F}\cdot\mathbf{r}. This is true for reasons I gave above. This is about as simple as you can get but only if you're thinking about work in mechanics.

    In conclusion, if you're looking for a nice visualisation of what the dot product is then you may not find one. But do you have a nice visualisation of what e.g. the cosine of an angle is? Then the next question to ask yourself is, "do I really need a nice visualisation of the dot product?”.
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    Amazingly explained, my man!
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    (Original post by Notnek)
    Why it's used in C4 should be clear - you can use it to find the angle between two vectors and also show that two vectors are perpendicular easily. It is defined as the product of the magnitudes divided by the cosine of the angle between the vectors.

    By "represent", I think you want to be able to visualise the dot product, just like you can visualise what the product of two numbers means. People try to give visualisations of the dot product but students don't usually like them because they're not immediately obvious - a product of two things should be simple they would say. But that's only because they're used to simple things like addition and multiplication of numbers. Why can't the dot product just be "product of the magnitudes divided by the cosine of the angle between the vectors"?

    Here's some geometric visualisations but you may not like them:

    If \mathbf{a} and \mathbf{b} are two vectors and \mathbf{b} is of unit length then \mathbf{a}\cdot \mathbf{b} is the projection of the vector \mathbf{a} on to \mathbf{b}. If you know mechanics, this means the component of \mathbf{a} in the direction of \mathbf{b}:





    But this only works if \mathbf{b} is a unit vector so you may think this is a bit crap. If \mathbf{b} is not a unit vector then \mathbf{a}\cdot \mathbf{b} is the projection of \mathbf{a} on to \mathbf{b} multiplied by the magnitude of \mathbf{b}. Nice? Not really.

    If you know about "work" in mechanics, this is the product of force (in direction of travel) and distance. If you have a force vector \mathbf{F} and a displacement vector \mathbf{r} then the work done by that force is \mathbf{F}\cdot\mathbf{r}. This is true for reasons I gave above. This is about as simple as you can get but only if you're thinking about work in mechanics.

    In conclusion, if you're looking for a nice visualisation of what the dot product is then you may not find one. But do you have a nice visualisation of what e.g. the cosine of an angle is? Then the next question to ask yourself is, "do I really need a nice visualisation of the dot product?”.
    Thanks. :hugs: I think I was looking for that nice visual representation that I now realise that I don't really need
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    (Original post by nyxnko_)
    Thanks. :hugs: I think I was looking for that nice visual representation that I now realise that I don't really need
    My mind manipulation worked! I should become a hypnotist.
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    (Original post by Notnek)
    My mind manipulation worked! I should become a hypnotist.
    wait... that was mind manipulation?!? :eek3: (maybe you should become a hypnotist instead of a tutor )
    tbf tho, you did clarify it quite well
 
 
 
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