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Vectors: direction vector - slope proof

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

(112)\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}

which is
=2(12)+(12) \frac{2}{\sqrt{(1^2)+(1^2)}}

=2 \sqrt{2}


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

(abc)\begin{pmatrix} a \\ b \\ c \end{pmatrix}

which is
=c(a2)+(b2) \frac{c}{\sqrt{(a^2)+(b^2)}}


^^ can someone show me the proof for this
(edited 13 years ago)
Reply 1
Avatar for nuodai
nuodai
17
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 yx\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, zx2+y2\dfrac{z}{\sqrt{x^2+y^2}}.

More generally, i.e. if you start from a point (x0,y0,z0)(x_0, y_0, z_0) rather than from the origin, this definition would give z−z0(x−x0)2+(y−y0)2\dfrac{z-z_0}{\sqrt{(x-x_0)^2 + (y-y_0)^2}}.
(edited 13 years ago)
Reply 2
Avatar for Milan.
Milan.
OP
3
Original post by nuodai

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 yx\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, zx2+y2\dfrac{z}{\sqrt{x^2+y^2}}.

More generally, i.e. if you start from a point (x0,y0,z0)(x_0, y_0, z_0) rather than from the origin, this definition would give z−z0(x−x0)2+(y−y0)2\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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