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Math Help - More Vector Questions

  1. #1
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    More Vector Questions

    Hi
    I need help on the following questions:
    1)Given a=i+2j+3k, b=2i-3j+2k and c=i+j-k, find the angle between b and c.
    This is what i have done: cos^{-1}\frac{2}{\sqrt17\sqrt1}
    =60'58'
    answers says its 114'50'

    2)Given x=2i-j+k and y=i+j+tk find the value of t if the angle between the two vectors is 60 degree

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following questions:
    1)Given a=i+2j+3k, b=2i-3j+2k and c=i+j-k, find the angle between b and c.
    This is what i have done: cos^{-1}\frac{2}{\sqrt17\sqrt1}
    =60'58'
    answers says its 114'50'

    2)Given x=2i-j+k and y=i+j+tk find the value of t if the angle between the two vectors is 60 degree

    P.S
    You have the wrong value for \mid c \mid

     \mid c \mid =\sqrt{1^2+1^2+(-1)^2}=\sqrt{3}

    Secondly, the dot product of  b\cdot c should be -3. So you need to find the inverse cosine cos^{-1}\frac{-3}{\sqrt17\sqrt3} which was around 114.84 degrees on my calculator.

    The dot product should have worked out like this:

    b \cdot c = (2)(1)+(-3)(1)+(2)(-1) =-3
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  3. #3
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    oops i c now
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  4. #4
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    could someone answer my second question?
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  5. #5
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    For the second problem, just start by writing what you know. The main equation you will need is:

    x \cdot y =\mid x\mid \mid y\mid cos(60^{\circ})

    The dot product is x \cdot y = 2-1+t=1+t

    cos(60^{\circ})=\frac{1}{2}

    So now the main equation becomes:

    1+t=\frac{\mid x\mid \mid y\mid}{2}

    You need the absolute values:

     \mid x \mid =2

    \mid y \mid =\sqrt{2+t^2}

    Substituting into the main equation results in:

    1+t=\sqrt{2+t^2}

    Square both sides to obtain:

    1+2t+t^2=2+t^2

    t=\frac{1}{2}
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  6. #6
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    ok thanks i'll tried that
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  7. #7
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    Quote Originally Posted by adkinsjr View Post
    For the second problem, just start by writing what you know. The main equation you will need is:

    x \cdot y =\mid x\mid \mid y\mid cos(60^{\circ})

    The dot product is x \cdot y = 2-1+t=1+t

    cos(60^{\circ})=\frac{1}{2}

    So now the main equation becomes:

    1+t=\frac{\mid x\mid \mid y\mid}{2}

    You need the absolute values:

     \mid x \mid =2

    \mid y \mid =\sqrt{2+t^2}

    Substituting into the main equation results in:

    1+t=\sqrt{2+t^2}

    Square both sides to obtain:

    1+2t+t^2=2+t^2

    t=\frac{1}{2}
    the book's answers says t=2 not a \frac{1}{2}
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  8. #8
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    Quote Originally Posted by Paymemoney View Post
    the book's answers says t=2 not a \frac{1}{2}
    Looking back, I see that I calculated \mid x \mid incorrectly. Sorry about that:

    \mid x\mid =\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}

    I originally had this as 2. So now we have to solve:

    1+t=\frac{\sqrt{6}\sqrt{2+t^2}}{2}

    The solution t=2 works for this. Sorry about that error.
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  9. #9
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    that's alright i get it now
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