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    So I've made a little start on this chapter but have found myself stuck already!!

    How do I do the last bit of this Example question after I have equated the coefficients?? I don't really understand how it goes from the 3rd from bottom line to the 2nd from bottom line of the workings

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    (Original post by Philip-flop)
    So I've made a little start on this chapter but have found myself stuck already!!

    How do I do the last bit of this Example question after I have equated the coefficients?? I don't really understand how it goes from the 3rd from bottom line to the 2nd from bottom line of the workings
    You equate coefficients in the same as as you'd equate coefficients of something like x^2+2x+1=Ax^2+Bx+C except here you are comparing the coefficients of the direction vectors \mathbf{a} and \mathbf{b}
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    (Original post by RDKGames)
    You equate coefficients in the same as as you'd equate coefficients of something like x^2+2x+1=Ax^2+Bx+C except here you are comparing the coefficients of the direction vectors \mathbf{a} and \mathbf{b}
    Yeah I think I understand the part where I compare coefficients but I dont really know the eliminating part which gives the 2nd from bottom line.

    Is it simply a case of using simultaneous equations where...
     -\land = (k-1) times both sides by -1 to give...  \land = (1-k)

    Sub this into...  (3k+1) = 4 \land
    which gives...  (3k+1) = 4(1-k)

    then solve for k from there?


    Sorry I'm not sure what the correct symbol is so I have used  \land instead
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    (Original post by Philip-flop)
    Yeah I think I understand the part where I compare coefficients but I dont really know the eliminating part which gives the 2nd from bottom line.

    Is it simply a case of using simultaneous equations where...
     -\land = (k-1) times both sides by -1 to give...  \land = (1-k)

    Sub this into...  (3k+1) = 4 \land
    which gives...  (3k+1) = 4(1-k)

    then solve for k from there?


    Sorry I'm not sure what the correct symbol is so I have used  \land instead
    Yeah thats what is being done here. Also the symbol is lambda.


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    If  k-1=-\lambda then  4\lambda = -4(k-1) by just multiplying both sides by 4.
    Now we have also that  3k+1=4\lambda so you can substitute  4\lambda with  -4(k-1) which gives  3k+1=-4(k-1) .
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    (Original post by RDKGames)
    Yeah thats what is being done here. Also the symbol is lambda.


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    Thank you!!

    So in a way, lambda  \lambda is used to describe scalars that are parallel and proportional to each other, right?

    (Original post by B_9710)
    If  k-1=-\lambda then  4\lambda = -4(k-1) by just multiplying both sides by 4.
    Now we have also that  3k+1=4\lambda so you can substitute  4\lambda with  -4(k-1) which gives  3k+1=-4(k-1) .
    Oh yeah. I didn't spot I could do it that way too! Thank you
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    (Original post by Philip-flop)
    Thank you!!

    So in a way, lambda  \lambda is used to describe scalars that are parallel and proportional to each other, right?
    Lambda is just a scalar variable which scales vectors by changing their magnitudes. It doesnt make sense when you say "used to describe scalars that are parallel and proportional" because scalars have no direction so how can they be parallel...?
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    (Original post by RDKGames)
    Lambda is just a scalar variable which scales vectors by changing their magnitudes. It doesnt make sense when you say "used to describe scalars that are parallel and proportional" because scalars have no direction so how can they be parallel...?
    Yeah your explanation makes more sense than mine! Thank you!! I had trouble putting it into words but had a brief understanding.

    So, lambda is a scalar variable which scales vectors by changing their magnitude. And the vectors have to be parallel too for them to have a related lambda? Sorry if I sound silly right now!
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    (Original post by Philip-flop)
    Yeah your explanation makes more sense than mine! Thank you!! I had trouble putting it into words but had a brief understanding.

    So, lambda is a scalar variable which scales vectors by changing their magnitude. And the vectors have to be parallel too for them to have a related lambda? Sorry if I sound silly right now!
    No, the scalar has nothing to do with dependency on two vectors being parallel or not, any vector can have its size changed by a vector regardless whether it is parallel to another vector or not. The only use for scalars here is that they change the magnitude, as previously mentioned, so you can use this to travel from some point A to some point B on whatever vector v by adding on \lambda \mathbf{v} onto the point A.
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    (Original post by Philip-flop)
    Yeah your explanation makes more sense than mine! Thank you!! I had trouble putting it into words but had a brief understanding.

    So, lambda is a scalar variable which scales vectors by changing their magnitude. And the vectors have to be parallel too for them to have a related lambda? Sorry if I sound silly right now!
    \lambda isn't anything special - it just represents a scalar variable.

    So if you have a vector \mathbf{a} then \lambda \mathbf{a} could be 2 \mathbf{a} or -3.1\boldsymbol{a} etc, depending on the value of \lambda.

    You can use any letter really as long as you don't underline it so it represents a scalar. \lambda and \mu are quite common.

    It is the case that a scalar \lambda multiplied by a vector \mathbf{a} will give you a new vector \lambda \mathbf{a}, which is parallel to \mathbf{a} but whose magnitude has changed by a factor of \lambda.
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    (Original post by RDKGames)
    No, the scalar has nothing to do with dependency on two vectors being parallel or not, any vector can have its size changed by a vector regardless whether it is parallel to another vector or not. The only use for scalars here is that they change the magnitude, as previously mentioned, so you can use this to travel from some point A to some point B on whatever vector v by adding on \lambda \mathbf{v} onto the point A.
    Ohhhh I see!! So lambda is literally a variable that just changes the magnitude of a vector. Thank yoooooou! Sorry for going round in circles a bit

    (Original post by notnek)
    \lambda isn't anything special - it just represents a scalar variable.

    So if you have a vector \mathbf{a} then \lambda \mathbf{a} could be 2 \mathbf{a} or -3.1\boldsymbol{a} etc, depending on the value of \lambda.

    You can use any letter really as long as you don't underline it so it represents a scalar. \lambda and \mu are quite common.

    It is the case that a scalar \lambda multiplied by a vector \mathbf{a} will give you a new vector \lambda \mathbf{a}, which is parallel to \mathbf{a} but whose magnitude has changed by a factor of \lambda.
    Wait so...  \lambda a will give a new vector that is parallel to the original vector  a ??
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    (Original post by Philip-flop)
    Wait so...  \lambda a will give a new vector that is parallel to the original vector  a ??
    Yes e.g. 2\mathbf{a} is parallel to \mathbf{a}:

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    (Original post by notnek)
    Yes e.g. 2\mathbf{a} is parallel to \mathbf{a}:

    Thanks notnek that made it so much easier to visualise
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    Stupid question of the day...
    What is a position vector?? Are these vectors that are from O the origin?

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    Like with this example. why is OP the position vector of P?
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    (Original post by Philip-flop)
    Stupid question of the day...
    What is a position vector?? Are these vectors that are from O the origin?

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    Like with this example. why is OP the position vector of P?
    Not a stupid question, and yes.

    The position vector of a point  P(x,y) is the vector from the origin  O to the point  P .
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    Can someone explain perpendicular and parallel vectors to me?...

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    I've read this little section multiple times but I don't think I completely understand. Obviously I know that if there are two lines perpendicular to each other that there will be an angle of 90 degrees, but the rest of it I'm not so sure about
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    (Original post by Philip-flop)
    Can someone explain perpendicular and parallel vectors to me?...

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    I've read this little section multiple times but I don't think I completely understand. Obviously I know that if there are two lines perpendicular to each other that there will be an angle of 90 degrees, but the rest of it I'm not so sure about
    Try writing out the dot product formula for parallel lines \theta = 0 and perpendicular lines \theta = 90.

    Then use the fact that \cos (0) = 1 and \cos (90) = 0.

    Please post your working if you're still unsure.
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    (Original post by Philip-flop)
    Can someone explain perpendicular and parallel vectors to me?...

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    I've read this little section multiple times but I don't think I completely understand. Obviously I know that if there are two lines perpendicular to each other that there will be an angle of 90 degrees, but the rest of it I'm not so sure about
    Firstly, are you happy about where and how we get \mathbf{a}\cdot \mathbf{b}=\lvert \mathbf{a}\lvert \lvert \mathbf{b} \lvert \cos(\theta) ?

    Secondly, if two vectors are parallel then the angle between them is always 0 so plugging this into the equation gives you the shown relation.

    Dunno but maybe watching this video can help since that relation is derived towards the end https://www.youtube.com/watch?v=WNuIhXo39_k
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    (Original post by notnek)
    Try writing out the dot product formula for parallel lines \theta = 0 and perpendicular lines \theta = 90.

    Then use the fact that \cos (0) = 1 and \cos (90) = 0.

    Please post your working if you're still unsure.
    Thank you notnek!! I understand perpendicular vectors a lot more now! But I'm still a bit "iffy" with parallel vectors.

    So for parallel vectors there will be no angle between them so if  \theta = 0^\circa then  cos(0 ^\circ ) =1

    because using the scalar product...
     cos(\theta) = \frac{a.b}{|a||b|}

     cos(0^\circ) = 1

    This shows that...
     \frac{a.b}{|a||b|} = 1 , when  \theta = 0^\circ

    Meaning...
     a.b = |a||b|

    (Original post by RDKGames)
    Firstly, are you happy about where and how we get \mathbf{a}\cdot \mathbf{b}=\lvert \mathbf{a}\lvert \lvert \mathbf{b} \lvert \cos(\theta) ?

    Secondly, if two vectors are parallel then the angle between them is always 0 so plugging this into the equation gives you the shown relation.

    Dunno but maybe watching this video can help since that relation is derived towards the end https://www.youtube.com/watch?v=WNuIhXo39_k
    I know the scalar product but proving how it's derived is another matter!

    I know that  a.b is the product of the x, y, and/or z coordinates of the vectors a and b
    and that  |a||b| is the magnitude/modulus/length of vectors a and b multiplied by each other.

    But as you can see my knowledge is still very limited.

    Thanks for the link, I will watch that video now
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    (Original post by Philip-flop)
    Thank you notnek!! I understand perpendicular vectors a lot more now! But I'm still a bit "iffy" with parallel vectors.

    So for parallel vectors there will be no angle between them so if  \theta = 0^\circa then  cos(0 ^\circ ) =1

    because using the scalar product...
     cos(\theta) = \frac{a.b}{|a||b|}

     cos(0^\circ) = 1

    This shows that...
     \frac{a.b}{|a||b|} = 1 , when  \theta = 0^\circ

    Meaning...
     a.b = |a||b|
    You say you are iffy but what you wrote is correct. Can you explain what you're unsure about?

    To summarise, the important thing is that you remember these:

    When \mathbf{a} and \mathbf{b} are perpendicular, \mathbf{a}\cdot \mathbf{b} = 0

    When \mathbf{a} and \mathbf{b} are parallel, \mathbf{a}\cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}|

    These results come from the scalar/dot product formula where \theta = 90 and \theta = 0

    For C4 you don't need to know how the scalar product formula is derived. You just need to be able to use the formula and use the results above that come from the formula.
 
 
 
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