C4 vectors help please

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  1. Baljit-Padda's Avatar
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    C4 vectors help please
    Hello! Ok so I am a bit confused. Why is it that sometimes when finding the angle between two lines you simply use the direction vectors (after the lamda or mew) as your a and b and sometimes you have to do like the vector AC dotted with AB and then the modulus's of those? How do you know when to use what?! :s :confused:
  2. Hobo389's Avatar
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    Re: C4 vectors help please
    Both are direction vectors, the AC or AB in your example and the direction vector within the vector equation of a line (I.e the second bit) that's why you use both. If you draw them it might help.
  3. raheem94's Avatar
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    Re: C4 vectors help please
    (Original post by Baljit-Padda)
    Hello! Ok so I am a bit confused. Why is it that sometimes when finding the angle between two lines you simply use the direction vectors (after the lamda or mew) as your a and b and sometimes you have to do like the vector AC dotted with AB and then the modulus's of those? How do you know when to use what?! :s :confused:
     \vec{AB} \text{ and } \vec{AC} also represent directions.
  4. the bear's Avatar
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    Re: C4 vectors help please
    sometimes you have to do like the vector AC dotted with AB

    this would be more if they asked you to find the angle between two vectors, rather than two lines...
  5. Baljit-Padda's Avatar
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    Re: C4 vectors help please
    OH! OK! silly me. So when we are asked to find the angle between two lines we would use the given direction vectors but if they asked for the angle between two vectors then you use the other
  6. Baljit-Padda's Avatar
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    Re: C4 vectors help please
    But take for example the june 2010 paper question 7b) It only asks for the angle ACB so I wouldn't know which method to use
  7. the bear's Avatar
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    Re: C4 vectors help please
    (Original post by Baljit-Padda)
    But take for example the june 2010 paper question 7b) It only asks for the angle ACB so I wouldn't know which method to use
    angle ACB is the angle between vectors CA and CB
  8. Baljit-Padda's Avatar
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    Re: C4 vectors help please
    I worked out AC (3,6,3) and BC (10,0,4) and then the modulus of AC is 3rt6 and the modulus of AB is 2rt29 I used the formula but come out with the wrong answer using these values.
  9. ztibor's Avatar
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    Re: C4 vectors help please
    (Original post by Baljit-Padda)
    Hello! Ok so I am a bit confused. Why is it that sometimes when finding the angle between two lines you simply use the direction vectors (after the lamda or mew) as your a and b and sometimes you have to do like the vector AC dotted with AB and then the modulus's of those? How do you know when to use what?! :s :confused:

    You can use any vectors being parallel to the lines.
    The simplest is to use the given direction vectors e.g. a and b,
    but if these are unknown you can use any vector 'representing' a
    line segment f.e. between points A and C, where A and C are on the
    same line. This vector AC maybe considered as direction vector too,
    (and using it in a given equation of a line as direction vector, you
    have to change only lamdba to an other parameter f.e beta to get the
    same vector for the running point of the line)

    To calculate the angle between two lines means to calculate it between
    the two direction vectors or between the two normal vector, wiches mutually perpendicular to the direction vectors so they have the same angle between each other than it does the direction vectors.
    For calculating the angle between two vectors we use the dot product od these vectors. THese product maybe calculated with two methods:
    1. From the known coordinates
    a(a1,a2,a3) and b(b1,b2,b3) -> a*b=a1*b1+a2*b2+a3*b3=>constant
    2. From the given modulus of the vectors and the angle between them
    a*b=|a|*|b|*cosA

    Taking equal the two equations
    cosA=\frac{\vec a \vdot \vec b}{|\vec{a}|\cdot |\vec{b}|}
    At the numerator you have to give the dot product from coordinates.
    Dividing it by the moduluses gives the cosine of angle.
    This means that we calculate the dot product of two unit-length vectors being parallel with the original vectors.
    cosA=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|}=\frac{\vec{a}}{|\vec{  a}|}\cdot \frac{\vec{b}}{|\vec{b}|}=\vec{u  }_a\cdot \vec{u}_b
  10. ztibor's Avatar
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    Re: C4 vectors help please
    (Original post by Baljit-Padda)
    I worked out AC (3,6,3) and BC (10,0,4) and then the modulus of AC is 3rt6 and the modulus of AB is 2rt29 I used the formula but come out with the wrong answer using these values.
    Did you get for angle 57.95 ?
  11. Baljit-Padda's Avatar
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    Re: C4 vectors help please
    (Original post by ztibor)
    You can use any vectors being parallel to the lines.
    The simplest is to use the given direction vectors e.g. a and b,
    but if these are unknown you can use any vector 'representing' a
    line segment f.e. between points A and C, where A and C are on the
    same line. This vector AC maybe considered as direction vector too,
    (and using it in a given equation of a line as direction vector, you
    have to change only lamdba to an other parameter f.e beta to get the
    same vector for the running point of the line)

    To calculate the angle between two lines means to calculate it between
    the two direction vectors or between the two normal vector, wiches mutually perpendicular to the direction vectors so they have the same angle between each other than it does the direction vectors.
    For calculating the angle between two vectors we use the dot product od these vectors. THese product maybe calculated with two methods:
    1. From the known coordinates
    a(a1,a2,a3) and b(b1,b2,b3) -> a*b=a1*b1+a2*b2+a3*b3=>constant
    2. From the given modulus of the vectors and the angle between them
    a*b=|a|*|b|*cosA

    Taking equal the two equations
    cosA=\frac{\vec a \vdot \vec b}{|\vec{a}|\cdot |\vec{b}|}
    At the numerator you have to give the dot product from coordinates.
    Dividing it by the moduluses gives the cosine of angle.
    This means that we calculate the dot product of two unit-length vectors being parallel with the original vectors.
    cosA=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|}=\frac{\vec{a}}{|\vec{  a}|}\cdot \frac{\vec{b}}{|\vec{b}|}=\vec{u  }_a\cdot \vec{u}_b
    Thank you SO much!!!
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