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Math Help - 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 \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.

    <br />
\bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}<br />

    How do you get that you have to do that:

    <br />
\bold{c}=|\bold{b}|\ \hat{\bold{a}}<br />

    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.

    <br />
\bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}<br />

    How do you get that you have to do that:

    <br />
\bold{c}=|\bold{b}|\ \hat{\bold{a}}<br />

    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.

    <br />
\bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}<br />

    How do you get that you have to do that:

    <br />
\bold{c}=|\bold{b}|\ \hat{\bold{a}}<br />

    Thanks so much.
    <br />
\bold{c}=|\bold{b}|\ \hat{\bold{a}}<br />

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

    <br />
||\bold{b}|\ \hat{\bold{a}}|=|\bold{b}||\hat{\bold{a}}|=|\bold{  b}|<br />

    RonL
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  9. #9
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    Hello, classicstrings!

    Let me give it a try . . .


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

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

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

    So now we know the length of \vec c.


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

    What is the magnitude of \vec a ?
    . . |\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 \sqrt{14} ? . . . Yes!

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

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

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

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

    This is the vector with the direction of \vec a and the length of \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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