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Thread: Vectors!

  1. #1
    Member classicstrings's Avatar
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    Vectors!

    Hello!

    If a = i - j + 5k and b = 2i - j - 3k find the vector c in the direction of a such that IcI = IbI -( magnitude)

    I would appreaciate a explanation to how to tackle this Q, so I can do it myself.

    Thanks!
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by classicstrings
    Hello!

    If a = i - j + 5k and b = 2i - j - 3k find the vector c in the direction of a such that IcI = IbI -( magnitude)

    I would appreaciate a explanation to how to tackle this Q, so I can do it myself.

    Thanks!
    You have the equation
    |c|=|b|
    you are given c is in direction of a
    c=ka, where k is any real no.

    Malay
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  3. #3
    Member classicstrings's Avatar
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    I still dont understand Malay.

    So Magnitude b = root(14)

    Then what?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by classicstrings
    Hello!

    If a = i - j + 5k and b = 2i - j - 3k find the vector c in the direction of a such that IcI = IbI -( magnitude)

    I would appreaciate a explanation to how to tackle this Q, so I can do it myself.

    Thanks!
    You seek $\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}$

    RonL
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  5. #5
    Member classicstrings's Avatar
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    Hi Capt Black,

    Using your formula i got [root(14)/3root(3)]*(i - j + 5k)

    But, I still do not understand why it is that.

    $\displaystyle
    \bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}
    $

    How do you get that you have to do that:

    $\displaystyle
    \bold{c}=|\bold{b}|\ \hat{\bold{a}}
    $

    Thanks so much.
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by classicstrings
    Hi Capt Black,

    Using your formula i got [root(14)/3root(3)]*(i - j + 5k)

    But, I still do not understand why it is that.

    $\displaystyle
    \bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}
    $

    How do you get that you have to do that:

    $\displaystyle
    \bold{c}=|\bold{b}|\ \hat{\bold{a}}
    $

    Thanks so much.
    Could you understand my equation?
    If yes, the aim is to determine k

    Keep Smiling
    Malay
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  7. #7
    Member classicstrings's Avatar
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    I'll go over it again tommorrow as it's 1:16 am here, and I need to get to school tommorrow! Thanks!
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by classicstrings
    Hi Capt Black,

    Using your formula i got [root(14)/3root(3)]*(i - j + 5k)

    But, I still do not understand why it is that.

    $\displaystyle
    \bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}
    $

    How do you get that you have to do that:

    $\displaystyle
    \bold{c}=|\bold{b}|\ \hat{\bold{a}}
    $

    Thanks so much.
    $\displaystyle
    \bold{c}=|\bold{b}|\ \hat{\bold{a}}
    $

    is a vector in the direction of $\displaystyle \bold{a}$ (that is it is a scalar times
    a unit vector pointing in the required direction. Also as $\displaystyle \hat{\bold{a}}$
    is a unit vector:

    $\displaystyle
    ||\bold{b}|\ \hat{\bold{a}}|=|\bold{b}||\hat{\bold{a}}|=|\bold{ b}|
    $

    RonL
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  9. #9
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    Hello, classicstrings!

    Let me give it a try . . .


    If $\displaystyle \vec a \,= \,i - j + 5k$ and $\displaystyle \vec b \,= \,2i - j - 3k$
    find the vector $\displaystyle \vec c$ in the direction of $\displaystyle \vec a$ such that $\displaystyle |\vec c| = |\vec b|$

    We want $\displaystyle \vec c$ to have the same magnitude as $\displaystyle \vec b$.

    Very well, let's find the magnitude (length) of $\displaystyle \vec b:$
    . . $\displaystyle |b|\:=\:\sqrt{2^2 + (-1)^2 + (-3)^2} \:= \:\sqrt{14}$

    So now we know the length of $\displaystyle \vec c$.


    We want $\displaystyle \vec c$ to have the same direction at $\displaystyle \vec a$.
    . . If $\displaystyle \vec a$ has the right magnitude $\displaystyle (\sqrt{14})$, we're done!

    What is the magnitude of $\displaystyle \vec a$ ?
    . . $\displaystyle |\vec a| \:= \:\sqrt{1^2 + (-1)^2 + 5&2} \:= \:\sqrt{27} \;= \;3\sqrt{3}$ . . . it's the wrong length

    Can we change its magnitude to$\displaystyle \sqrt{14}$ ? . . . Yes!

    If we divide $\displaystyle \vec a$ by $\displaystyle 3\sqrt{3}$, we have: .$\displaystyle \vec u \:=\:\frac{\vec a}{3\sqrt{3}}$

    . . This is a unit vector in the direction of $\displaystyle \vec a$ (exactly one unit long).

    To make it$\displaystyle \sqrt{14}$ units long, multiply it by$\displaystyle \sqrt{14}$, and we have: .$\displaystyle \sqrt{14}\cdot\frac{\vec a}{3\sqrt{3}} $

    So we have: .$\displaystyle \vec c \;\;=\; \;\frac{\sqrt{14}}{3\sqrt{3}}\cdot\vec a \quad\Rightarrow$ . $\displaystyle \boxed{\vec c\;= \;\frac{\sqrt{42}}{9}(i - j + 5k)}$

    This is the vector with the direction of $\displaystyle \vec a$ and the length of $\displaystyle \vec b$.

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  10. #10
    Super Member malaygoel's Avatar
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    Quote Originally Posted by classicstrings
    Hello!

    If a = i - j + 5k and b = 2i - j - 3k find the vector c in the direction of a such that IcI = IbI -( magnitude)

    I would appreaciate a explanation to how to tackle this Q, so I can do it myself.

    Thanks!
    I would like to tell you something about vectors.
    A vector has a direction and magnitude.These two properties are independent of each other. A vector is completely determined by its direction and magnitude.
    There are two ways of representating a vector:
    1)Using Cartesian co-ordinates
    2)Using Polar co-ordinates
    (These can be interchanged from one form to another, use the form you are comfortable with)
    If two vectors have the same direction, then one can be expressed as a scalar times another.
    In your question, c has a given magnitude(equal to b), and a definite direction(same as a). So, your vector c is unique.
    (I think it may be help your revision, I will post more later)

    Keep Smiling
    Malay
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