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Math Help - vectors

  1. #1
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    vectors

    The points A, B and C have position vectors a = ( 2, 1, 2) b= (-3, 2, 5)
    c= ( 4, 5, -2) respectively with respect to a fixed origin. The point D is such that ABCD, in that order is a parallelogram.

    a) Find the position vector of D

    b) Fİnd the position vector of the point at which the diagonals of the parallelogram interesct

    c) Calculate the angle BAC, giving your answer to the nearst tenth of a degree.

    Thank you in advance! I would really appreciate it if you showed the steps.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Oasis1993 View Post
    The points A, B and C have position vectors a = ( 2, 1, 2) b= (-3, 2, 5)
    c= ( 4, 5, -2) respectively with respect to a fixed origin. The point D is such that ABCD, in that order is a parallelogram.

    a) Find the position vector of D

    b) Fİnd the position vector of the point at which the diagonals of the parallelogram interesct

    c) Calculate the angle BAC, giving your answer to the nearst tenth of a degree.

    Thank you in advance! I would really appreciate it if you showed the steps.
    a) Let the position vector of D be (x, y, z).

    Midpoint of AC=Midpoint of BD

    (2+4, 1+5, 2-2)=(-3+x, 2+y, 5+z)

    (x, y, z)=(9, 4, -5)

    b) The point at which the diagonals of the parallelogram intersect=Midpoint of AC= (\frac{2+4}{2}, \frac{1+5}{2}, \frac{2-2}{2})=(3, 3, 0)
    Last edited by alexmahone; January 13th 2010 at 09:27 AM. Reason: Typo
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  3. #3
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    Hello Oasis1993
    Quote Originally Posted by Oasis1993 View Post
    The points A, B and C have position vectors a = ( 2, 1, 2) b= (-3, 2, 5)
    c= ( 4, 5, -2) respectively with respect to a fixed origin. The point D is such that ABCD, in that order is a parallelogram.

    a) Find the position vector of D

    b) Fİnd the position vector of the point at which the diagonals of the parallelogram interesct

    c) Calculate the angle BAC, giving your answer to the nearst tenth of a degree.

    Thank you in advance! I would really appreciate it if you showed the steps.
    I agree with alexmahone's answer to (b), but there's a simple typo in (a):
    Quote Originally Posted by alexmahone View Post
    ...

    (2+4, 1+5, 2-2)=(-3+x, 2+y, \color{red}3\color{black}+z)

    (x, y, z)=(9, 4, \color{red}-3\color{black})
    ...
    It should be (9,4,-5).

    Here's the outline for part (c). Use the dot (scalar) product formula:
    \vec{AB}.\vec{AC} = |AB||AC|\cos(\angle BAC)

    \Rightarrow\cos(\angle BAC)=\frac{\vec{AB}.\vec{AC}}{|AB||AC|}
    Now
    \vec{AB} = \vec{b}-\vec{a}=\begin{pmatrix}-3\\2\\5\end{pmatrix}-\begin{pmatrix}2\\1\\2\end{pmatrix}=\begin{pmatrix  }-5\\1\\3\end{pmatrix}
    and
    \vec{AC} = \vec{c}-\vec{a}=...
    Also
    |AB|=\sqrt{(-5)^2+1^2+3^2}= \sqrt{35}
    and
    |AC| = ...
    The dot product is:
    \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} .\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}=x_1x_2+  y_1y_2+z_1z_2
    So
    \vec{AB}.\vec{AC} = ...
    Can you complete it now?

    Grandad
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