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  1. #1
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    equations planes

    equations planes-4-8-1.jpg
    equations planes-4-8-2.jpg
    equations planes-4-8-3.jpg
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  2. #2
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    Quote Originally Posted by Raiden_11 View Post
    ...
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    Hello,

    maybe this comes a little bit late ...
    If \vec{x}=\left(\begin{array}{c}x \\ y \\z \end{array} \right) then the equations of the lines can be written like this: L:\vec{x} = \underbrace{\vec{a}}_{\text{fixed point}} + r \cdot \underbrace{\vec{u}}_{\text{direction vector}}
    L_1: \vec{x} = \left(\begin{array}{c}5 \\ -1 \\1 \end{array} \right) + k \cdot \left(\begin{array}{c}3 \\ -4 \\0 \end{array} \right) and

    L_2: \vec{x} = \left(\begin{array}{c}1 \\ 1 \\1 \end{array} \right) + t \cdot \left(\begin{array}{c}1 \\ 2 \\-2 \end{array} \right)

    If \vec{a} and \vec{b} are the position vectors to the fixed points of the lines, \vec{u} and \vec{v} the direction vectors of the lines and \vec{n} is the normal vector to \vec{u} and \vec{v} then the distance d is calculated by:

    d = \frac{(\vec{b} - \vec{a}) \cdot \vec{n}}{|\vec{n}|} (see attachment)

    1. \vec{n} = \left(\begin{array}{c}3 \\ -4 \\0 \end{array} \right) \times \left(\begin{array}{c}1 \\ 2 \\-2 \end{array} \right) = \left(\begin{array}{c}8 \\ 6 \\10 \end{array} \right) thus \left| \left(\begin{array}{c}8 \\ 6 \\10 \end{array} \right) \right| = \sqrt{200} = 10 \cdot \sqrt{2}

    2. Now plug in all the values you know into the equation to calculate the distance:

    d = \frac{\left( \left(\begin{array}{c}1 \\ 1 \\1 \end{array} \right) - \left(\begin{array}{c}5 \\ -1 \\1 \end{array} \right) \right) \cdot \left(\begin{array}{c}8 \\ 6 \\10 \end{array} \right)}{10 \cdot \sqrt{2}} = \frac{-20}{10 \cdot \sqrt{2}} = -\sqrt{2}
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  3. #3
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    Quote Originally Posted by Raiden_11 View Post
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    Hello,

    let \overrightarrow{n_i} the normal vector of the plane p_i that means \overrightarrow{n_1} is :
    \overrightarrow{n_1} = \left( \begin{array}{c}a_1 \\ b_1 \\c_1 \end{array}\right)

    to a) \overrightarrow{n_1} = k \cdot \overrightarrow{n_2} = t \cdot \overrightarrow{n_3} and d_1 \ne d_2 \wedge d_1 \ne d_3 \wedge d_2 \ne d_3

    to b) \overrightarrow{n_1} = k \cdot \overrightarrow{n_2} \text{ and } \overrightarrow{n_1} \ne  \overrightarrow{n_3} \text{ and } \overrightarrow{n_2} \ne  \overrightarrow{n_3}

    to c) \overrightarrow{n_1} \ne  \overrightarrow{n_2} \text{ and }  \overrightarrow{n_3} =  k \cdot \overrightarrow{n_1} + t \cdot \overrightarrow{n_2} \text{ and } d_3 = k \cdot d_1 + t \cdot d_2

    to d) \overrightarrow{n_1} \times  \overrightarrow{n_2} \text{ and } \overrightarrow{n_1} \times  \overrightarrow{n_3} \text{ and } d_1 \ne d_2 \wedge d_1 \ne d_3 \wedge d_2 \ne d_3

    to e) That's the usual situation: No numerical dependencies(?) between the normal vectors and the constants.
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  4. #4
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    what about the third one?

    Thanks for the help earlier can u helpme solve this one too.

    equations planes-4-8-3.jpg
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  5. #5
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    Quote Originally Posted by Raiden_11 View Post
    Thanks for the help earlier can u helpme solve this one too.

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    Hello,

    to a): I assume that you have written your problem correctly.
    If so you can simplify the given equations:
    x = (2-k)(4)-3k = 8-7k
    y = (2-k)(-6) = -12 + 6k
    z = (2-k)+2k = 2 + k

    If \vec{x}=\left(\begin{array}{c}x\\y\\z \end{array}\right)

    Then these equations with the given restriction describe a straight line connecting the points A(8, -12, 2) [for k = 0] and B(-6, 0, 4) [for k = 2]:

    The equation of the line where the points A and B are placed on is:

    \vec{x}=\left(\begin{array}{c}8\\ -12 \\ 2 \end{array}\right) + k \cdot \left(\begin{array}{c}-7 \\ 6\\ 1 \end{array}\right)

    to b) I have no idea what to do - sorry!
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