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    vecctor- line parallel / in plane

    for part b and part c, i shown my working in the photo below. The ans is just below the question . pease refer it. i cant understand why the ans explain. can anyone explain in a more detailed way ?vecctor- line parallel / in plane-dsc_0003-2-1-.jpgvecctor- line parallel / in plane-dsc_0001-9-1-.jpgvecctor- line parallel / in plane-dsc_0002-5-1-.jpg
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    Re: vecctor- line parallel / in plane

    Quote Originally Posted by delso View Post
    for part b and part c, i shown my working in the photo below. The ans is just below the question . pease refer it. i cant understand why the ans explain. can anyone explain in a more detailed way ?Click image for larger version. 

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    Good morning,

    I'll do the 1st example and leave the rest for you:

    Equation of the plane: p: \begin{pmatrix}2\\1\\-1\end{pmatrix} \cdot \vec x = 3

    Equation of the line: l: \vec r = \begin{pmatrix}1\\6\\-1\end{pmatrix} + \lambda \cdot \begin{pmatrix}-1\\2\\-3\end{pmatrix}

    To determine the points of intersection replace \vec x in p by \vec r of l and solve for \lambda:

    \begin{pmatrix}2\\1\\-1\end{pmatrix} \cdot \left( \begin{pmatrix}1\\6\\-1\end{pmatrix} + \lambda \cdot \begin{pmatrix}-1\\2\\-3\end{pmatrix} \right) = 3~\implies~9+3 \lambda = 3~\implies~ \lambda = -2

    Replace \lambda by (-2) in \vec r and you'll get \begin{pmatrix}3\\2\\5\end{pmatrix}

    To do the following examples you should know which value of \lambda you'll get if
    • the line is parallel to the plane (no point of intersection!)
    • the line lies in the plane (unlimited number of points of intersection)
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    Re: vecctor- line parallel / in plane

    Quote Originally Posted by earboth View Post
    Good morning,

    I'll do the 1st example and leave the rest for you:

    Equation of the plane: p: \begin{pmatrix}2\\1\\-1\end{pmatrix} \cdot \vec x = 3


    Equation of the line: l: \vec r = \begin{pmatrix}1\\6\\-1\end{pmatrix} + \lambda \cdot \begin{pmatrix}-1\\2\\-3\end{pmatrix}

    To determine the points of intersection replace \vec x in p by \vec r of l and solve for \lambda:

    \begin{pmatrix}2\\1\\-1\end{pmatrix} \cdot \left( \begin{pmatrix}1\\6\\-1\end{pmatrix} + \lambda \cdot \begin{pmatrix}-1\\2\\-3\end{pmatrix} \right) = 3~\implies~9+3 \lambda = 3~\implies~ \lambda = -2

    Replace \lambda by (-2) in \vec r and you'll get \begin{pmatrix}3\\2\\5\end{pmatrix}

    To do the following examples you should know which value of \lambda you'll get if
    • the line is parallel to the plane (no point of intersection!)
    • the line lies in the plane (unlimited number of points of intersection)
    well, i have done the part b and c in the previous poat. how can the ans relate to the line is parallel to the plane (no point of intersection!) and
    the line lies in the plane (unlimited number of points of intersection ?
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    Re: vecctor- line parallel / in plane

    Quote Originally Posted by delso View Post
    well, i have done the part b and c in the previous poat. how can the ans relate to the line is parallel to the plane (no point of intersection!) and
    the line lies in the plane (unlimited number of points of intersection ?
    In my opinion you haven't done these questions

    You are asked to solve equations but you have scribbled down some calculations. To part b):

    The final line of the equation should read
    \underbrace{-2}_{\text{your result}} = 3
    So which value has \lambda? What does that mean if you are looking for points of intersections?

    To part c):

    The final line of the equation should read
    \underbrace{3}_{\text{your result}} = 3
    So which value has \lambda? What does that mean if you are looking for points of intersections?
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    Re: vecctor- line parallel / in plane

    so part part c has \lambda . part b has no \lambda. so for the equation has no \lambda means the line is parallel? the line has \lambda=1 means the line has unlimited point of intersection. ? well, i cant understand why \lambda = 1, the equation has unlimited point of intersection?
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    Re: vecctor- line parallel / in plane

    Quote Originally Posted by delso View Post
    so part part c has \lambda . part b has no \lambda. so for the equation has no \lambda means the line is parallel? the line has \lambda=1 means the line has unlimited point of intersection. ? well, i cant understand why \lambda = 1, the equation has unlimited point of intersection?
    Good morning,
    I guess that you do such excercises in school. Would your teacher accept your explanation as a correct one? Hmmm?

    The final line of the equation should read
    \underbrace{-2}_{\text{your result}} = 3
    So which value has \lambda? What does that mean if you are looking for points of intersections?
    This equation is false. There doesn't exist a value of \lambda which will turn this equation to be true. So if there isn't an appropriate \lambda then there isn't any point of intersection. Therefore the line and the plane must be parallel.


    Now you use my argumentation to show that in part c) you can take every real number for \lambda. And then you have to deduce what this means geometrically.
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