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Math Help - Vector Questions

  1. #1
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    Vector Questions

    a = 3i + 2j b=-9i + t. Determine t

    Angle between vector a and b is 60

    Given the vectors u = (-3,0,1) v= (-1,-2,2) determine

    unit vector in the direction of v

    v x u

    Find the value of the direction angles of both vectors
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  2. #2
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    Quote Originally Posted by peon123 View Post
    a = 3i + 2j b=-9i + t. Determine t

    Angle between vector a and b is 60
    The formula for the angle between two vectors a & b is
    \cos \theta = \frac{a \cdot b}{|a||b|}.

    \cos 60 = \frac{3(-9) + 2t}{\sqrt{3^2 + 2^2}\sqrt{(-9)^2 + t^2}}

    \frac{1}{2} = \frac{-27 + 2t}{\sqrt{13}\sqrt{81 + t^2}}

    \frac{1}{2} = \frac{-27 + 2t}{\sqrt{1053 + 13t^2}}

    \sqrt{1053 + 13t^2} = 2(-27 + 2t)

    \sqrt{13t^2 + 1053} = 4t - 54

    13t^2 + 1053 = 16t^2 - 432t + 2916

    0 = 3t^2 - 432t + 1863

    0 = 3(t^2 - 144t + 621)

    0 = t^2 - 144t + 621

    t = \frac{144 \pm \sqrt{(-144)^2 - 4(1)(621)}}{2(1)}

    t = \frac{144 \pm \sqrt{18252}}{2}

    t = \frac{144 \pm 78\sqrt{3}}{2}

    t = 72 \pm 39\sqrt{3}

    So t = 4.450 or 139.550, but it looks like t = 4.450 is extraneous (the angle turns out to be 120 degrees).
    I think that's right.


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  3. #3
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    Quote Originally Posted by peon123 View Post
    Given the vectors u = (-3,0,1) v= (-1,-2,2) determine

    unit vector in the direction of v
    Be careful with notation. For vectors, use "<>", not parentheses.
    v = <-1, -2, 2>
    unit vector in the direction of v = \frac{v}{|v|}

    \frac{<-1, -2, 2>}{\sqrt{(-1)^2 + (-2)^2 + 2^2}}

    \frac{<-1, -2, 2>}{\sqrt{9}}

    \frac{<-1, -2, 2>}{3}

    \left\langle -\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right\rangle


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