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Math Help - Vertices help

  1. #1
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    Vertices help

    a) A triangle ABC has vertices given by A(1,1,8), B(4,-3,-4) and C(-3,1,5)

    (i) Determine the distance between B and C (in units).
    (ii) Find the angle formed at the vertex A (2 decimal places)


    Please help...
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  2. #2
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    Hello, Unt0t!

    A triangle ABC has vertices given by A(1,1,8), B(4,-3,-4) and C(-3,1,5)

    (a) Determine the distance between B and C (in units).
    You don't know the Distance Formula?

    BC \:=\:\sqrt{(-3-4)^2 + (1-[-3])^2 + (5-[-4])^2} \:=\:\sqrt{49 + 16 + 81} \:=\:\sqrt{146}



    (b) Find the angle formed at the vertex A (2 decimal places)
    The angle \theta between two vectors, \vec{u},\:\vec{v}, is given by: . \cos\theta\;=\;\frac{|\vec{u}\cdot\vec{v}|}{|\vec{  u}||\vec{v}|}

    Let: \vec{u} \:=\:AB \:=\:\langle4-1,\:\text{-}3-1,\:\text{-}4-8\rangle \:=\:\langle3,\,\text{-}4,\,\text{-}12\rangle

    Let: \vec{v} \:=\:AC\:=\:\langle\text{-}3-1,\:1-1,\:5-8\rangle \:=\:\langle\text{-}4,\:0,\:\text{-}3\rangle


    Then: . \cos A \:=\:\frac{|\langle 3,\,\text{-}4,\,\text{-}12\rangle\cdot\langle\text{-}4,\,0,\,\text{-}3\rangle|}{\sqrt{3^2+(\text{-}4)^2+(\text{-}12)^2}\,\sqrt{(\text{-}4)^2+0^2 + (\text{-}3)^2}} \:=\:\frac{|(3)(\text{-}4) + (\text{-}4)(0) + (\text{-}12)(\text{-}3)|}{\sqrt{9 + 16 + 144}\,\sqrt{16 + 0 + 9}}

    . . . . . . . . = \;\frac{-12 + 0 + 36}{\sqrt{169}\,\sqrt{25}} \;=\;\frac{24}{13\cdot5}


    Therefore: . A \;=\;\cos^{-1}\left(\frac{24}{65}\right) \;\approx\;68.33^o

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  3. #3
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    Thanks, i've got another question now.

    Let u = (1, -3, -1), v = (1, 1, 2), w = (-2, -2, 1), x = (-1, -1, 1).

    a) By considering the dot product only, identify 2 vectors that will satisfy each of the following situations:

    (i) the angle between them is less than (pie)/2;
    (ii) the angle between them is greater than (pie)/2;

    Justify your answers by considering the relationship between the dot product and the cosine of the angle.

    b) Identify 2 vectors that will be perpendicular to each other. Justify your answer.
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  4. #4
    MHF Contributor red_dog's Avatar
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    If \overrightarrow{u}=(x_1,y_1,z_1) and \overrightarrow{v}=(x_2,y_2,z_2) then \overrightarrow{u}\cdot\overrightarrow{v}=x_1x_2+y  _1y_2+z_1z_2

    We have
    \overrightarrow{u}\cdot\overrightarrow{v}>0\Leftri  ghtarrow 0\leq\widehat{\overrightarrow{u}\overrightarrow{v}  }<\frac{\pi}{2}
    \overrightarrow{u}\cdot\overrightarrow{v}<0\Leftri  ghtarrow \frac{\pi}{2}<\widehat{\overrightarrow{u}\overrigh  tarrow{v}}\leq\pi
    \overrightarrow{u}\cdot\overrightarrow{v}=0\Leftri  ghtarrow \widehat{\overrightarrow{u}\overrightarrow{v}}=\fr  ac{\pi}{2}.
    Now, using these, you can solve the problem.
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