Zoran Skoda sb 3

t(a,b)R 3 t\in (a,b)\mapsto \mathbf{R}^3
r(t)=r 0+ta \vec{r}(t) = \vec{r}_0 + t \vec{a}
x(t)=x 0+a xt y(t)=y 0+a yt z(t)=z 0+a zt\array{ x(t) = x_0 + a_x t \\ y(t) = y_0 + a_y t \\ z(t) = z_0 + a_z t }
x(t)=2+3t y(t)=1+t z(t)=3t\array{ x(t) = 2 + 3t \\ y(t) = 1 + t \\ z(t) = 3 - t }

t = 0, A(2,1,3), B(5,2,2)

Isključimo varijablu t.

x = 2 + 3 t

3 t = x -2

t = x/3 -2/3

y = 1 + x/3 -2/3 = x/3 - 2/3

z = 3 - (x/3 - 2/3) = -x/3 + 7/3

R0(2,1,2)

R1(3,2,4)

R2(0,2,-1)

vector(R0,R1) = (1,1,2)

vector(R0,R2) = (-2,1,-3)

vector r(t) = (2,1,2)+u(1,1,2)+v(-2,1,-3)

   r(t) = (2+u-2v, 1+u+v,2+2u-3v)

Last revised on November 16, 2020 at 18:35:37. See the history of this page for a list of all contributions to it.