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Vectors: direction vector - slope proof watch

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    write down the slope of the line in the direction of the vector

    \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}

    which is
    = \frac{2}{\sqrt{(1^2)+(1^2)}}

    = \sqrt{2}


    and the general form
    write down the slope of the line in the direction of the vector

    \begin{pmatrix} a \\ b \\ c \end{pmatrix}

    which is
    = \frac{c}{\sqrt{(a^2)+(b^2)}}


    ^^ can someone show me the proof for this
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    By the looks of it it's just a definition; but intuitively, you generalise from the 2D case.

    The slope is how far you travelled 'up' divided by how much you travelled 'across'. In the 2D case this simply means \dfrac{y}{x}, but in the 3D case, 'up' is now the z-direction, and 'across' is how far along the (x,y)-plane you travelled. Hence, \dfrac{z}{\sqrt{x^2+y^2}}.

    More generally, i.e. if you start from a point (x_0, y_0, z_0) rather than from the origin, this definition would give \dfrac{z-z_0}{\sqrt{(x-x_0)^2 + (y-y_0)^2}}.
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    (Original post by nuodai)
    By the looks of it it's just a definition; but intuitively, you generalise from the 2D case.

    The slope is how far you travelled 'up' divided by how much you travelled 'across'. In the 2D case this simply means \dfrac{y}{x}, but in the 3D case, 'up' is now the z-direction, and 'across' is how far along the (x,y)-plane you travelled. Hence, \dfrac{z}{\sqrt{x^2+y^2}}.

    More generally, i.e. if you start from a point (x_0, y_0, z_0) rather than from the origin, this definition would give \dfrac{z-z_0}{\sqrt{(x-x_0)^2 + (y-y_0)^2}}.
    thank you!
    just needed someone to point out that bit of intuition...
    too much stats in the past week has ruined me
 
 
 
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