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Math Help - Vector Problem

  1. #1
    Junior Member Morphayne's Avatar
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    Exclamation Vector Problem

    a = vector a
    b = vector b
    [ ] = numbers in squared brackets are vector components
    |a - b| = the magnitude of vector a minus vector b

    Problem:

    If a=[k, 2] and b=[7,6], where k is a real number, determine all values of k such that |a - b| = 5.
    The answer in the book says, "10 or 4."


    Please help me I have a test tomorrow and I desperately need help on this question.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Morphayne View Post
    a = vector a
    b = vector b
    [ ] = numbers in squared brackets are vector components
    |a - b| = the magnitude of vector a minus vector b

    Problem:

    If a=[k, 2] and b=[7,6], where k is a real number, determine all values of k such that |a - b| = 5.


    Please help me I have a test tomorrow and I desperately need help on this question.
    Lets start by doing the computations

    |(k,2)-(7,6)|=|(k-7,-4)|

    Now lets compute the magnitude

    |(k-7),-4|=\sqrt{(k-7)^2+(-4)^2}

    Now we need to set the above equal to 5 and solve.

    \sqrt{(k-7)^2+(-4)^2}=5

    Just solve from here
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  3. #3
    Junior Member Morphayne's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Lets start by doing the computations

    |(k,2)-(7,6)|=|(k-7,-4)|

    Now lets compute the magnitude

    |(k-7),-4|=\sqrt{(k-7)^2+(-4)^2}

    Now we need to set the above equal to 5 and solve.

    \sqrt{(k-7)^2+(-4)^2}=5

    Just solve from here
    Do you mind solving it? I'm a struggling student. I gave you +Thanks.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Morphayne View Post
    a = vector a
    b = vector b
    [ ] = numbers in squared brackets are vector components
    |a - b| = the magnitude of vector a minus vector b

    Problem:

    If a=[k, 2] and b=[7,6], where k is a real number, determine all values of k such that |a - b| = 5.


    Please help me I have a test tomorrow and I desperately need help on this question.
    If \bold{a}=<k,2> and \bold{b}=<7,6>, then \bold{a}-\bold{b}=<k-7,2-6>=<k-7,-4>. Therefore:

    {\left|\bold{a}-\bold{b}\right|}=\sqrt{(k-7)^2+(-4)^2}=\sqrt{k^2-14k+49+16}=\sqrt{k^2-14k+65}

    Now find where
    {\left|\bold{a}-\bold{b}\right|}=5.

    5=\sqrt{k^2-14k+65}
    \rightarrow 25=k^2-14k+65
    \rightarrow k^2-14k+40=0
    (k-4)(k-10)=0
    \rightarrow k=4 or k=10.

    You can plug these back in to verify the k values.


    5=\sqrt{4^2-14(4)+65}=\sqrt{16-56+65}=\sqrt{25}={\color{red}5}
    5=\sqrt{10^2-14(10)+65}=\sqrt{100-140+65}=\sqrt{25}={\color{red}5}

    Hope that answers your question!!
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  5. #5
    Junior Member Morphayne's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    If \bold{a}=<k,2> and \bold{b}=<7,6>, then \bold{a}-\bold{b}=<k-7,2-6>=<k-7,-4>. Therefore:

    {\left|\bold{a}-\bold{b}\right|}=\sqrt{(k-7)^2+(-4)^2}=\sqrt{k^2-14k+49+16}=\sqrt{k^2-14k+65}

    Now find where
    {\left|\bold{a}-\bold{b}\right|}=5.

    5=\sqrt{k^2-14k+65}
    \rightarrow 25=k^2-14k+65
    \rightarrow k^2-14k+40=0
    (k-4)(k-10)=0
    \rightarrow k=4 or k=10.

    You can plug these back in to verify the k values.


    5=\sqrt{4^2-14(4)+65}=\sqrt{16-56+65}=\sqrt{25}={\color{red}5}
    5=\sqrt{10^2-14(10)+65}=\sqrt{100-140+65}=\sqrt{25}={\color{red}5}

    Hope that answers your question!!
    THANK YOU SO MUCH!
    +THANKS!!
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