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Math Help - Vectors

  1. #1
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    Vectors

    I have always had problems with vectors and now need help to understend this simpe task:

    Given triangle ABC with vertices A(-2; 5), B(-5; 4), C(1;1).

    Find:
    1)vecto \overrightarrow{AB} \ \overrightarrow{CB}\ \overrightarrow{CA} coordinates and length

    2) internal angle to vertice B

    3) general equation of line AB

    4) general equation of height CD

    5)edge AB and height CD intersection D

    For 1st I think that  \overrightarrow{AB}=\sqrt{(-5+2)^2+(4-5)^2}=\sqrt{10}

    \overrightarrow{CB}=\sqrt{(-5-1)^2+(4-1)^2}=\sqrt{45}

    \overrightarrow{CA}=\sqrt {(-2-1)^2+(-1)^2}=5
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  2. #2
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    Your notation is incorrect.
    \begin{gathered}  \overrightarrow {AB}  =  - 3i - j = \left\langle { - 3, - 1} \right\rangle  \hfill \\  \left\| {\overrightarrow {AB} } \right\| = \left\| {\left\langle { - 3, - 1} \right\rangle } \right\| = \sqrt {\left( { - 3} \right)^3  + \left( { - 1} \right)^2 }  \hfill \\ \end{gathered}

    Then \overrightarrow {AC}  = \left\langle {3, - 4} \right\rangle \; \Rightarrow \;m\left( {\angle BAC} \right) = \arccos \left( {\frac{{\overrightarrow {AB}  \cdot \overrightarrow {AC} }}<br />
{{\left\| {\overrightarrow {AB} } \right\|\left\| {\overrightarrow {AC} } \right\|}}} \right)

    Now you do the others.
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  3. #3
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    Quote Originally Posted by Bernice View Post
    I have always had problems with vectors and now need help to understend this simpe task:

    Given triangle ABC with vertices A(-2; 5), B(-5; 4), C(1;1).

    Find:
    ...

    3) general equation of line AB

    4) general equation of height CD

    5)edge AB and height CD intersection D

    For 1st I think that  \overrightarrow{AB}=\sqrt{(-5+2)^2+(4-5)^2}=\sqrt{10}

    \overrightarrow{CB}=\sqrt{(-5-1)^2+(4-1)^2}=\sqrt{45}

    \overrightarrow{CA}=\sqrt {(-2-1)^2+(-1)^2}=5
    to #3: There are 4 different equations:

    \overrightarrow{r(t)}=(-2,5)+t \cdot ((-5,4) - (-2,5))=(-2,5)+t \cdot (-3,-1)
    These equations describe the same line:
    Passing through B in direction AB
    Passing through A in direction BA
    Passing through B in direction BA

    to #4: The height CD is perpendicular to AB, that means \overrightarrow{CD} is a normal vector to \overrightarrow{AB}:

    \overrightarrow{AB} = (-3,-1)~\implies~\overrightarrow{n_{AB}}= (1,-3) . Thus the equation of CD is:


    \overrightarrow{r(s)}=(1,1)+s \cdot (1,-3)

    to #5: Calculate the coordinates of the intersection point of the line in #3 and in #4:
    \overrightarrow{r(t)} = \overrightarrow{r(s)} will yield a system of simultaneous equations:
    (-2,5)+t \cdot (-3,-1) = (1,1)+s \cdot (1,-3)
    t \cdot (-3,-1) - s \cdot (1,-3)= (1,1)-(-2,5)
    t \cdot (-3,-1) - s \cdot (1,-3)= (3,-4)

    \left|\begin{array}{rcl}-3t-s&=&3\\-t+3s&=&-4\end{array}\right.

    I've got D\left(-\dfrac12\ ,\ \dfrac{11}2\right)
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