Question

find a vector parametrization v(t),t (0,3) of the path(or loop) ABCA where A(1,0,1),B(1,1,0),C(0,1,1)

Answer 1

Answer on Question #51514 – Math – Differential Geometry

find a vector parametrization $v(t), t(0,3)$ of the path (or loop) ABCA where $A(1,0,1), B(1,1,0), C(0,1,1)$

Solution

Path $ABCA$ consists of lines $AB, BC, CA$.

Apply the parametric form of the equation of a line $x = x_0 + at, y = y_0 + bt, z = z_0 + ct$, where $(x, y, z)$ is any point on the line, $(x_0, y_0, z_0)$ is a fixed point on the line and $\vec{l} = \langle a, b, c \rangle$ is some vector that is parallel to the line.

Line $AB$ has $(x_0, y_0, z_0) = (x_A, y_A, z_A) = (1, 0, 1)$,

$$\vec{l} = \overrightarrow{AB} = \langle x_B - x_A, y_B - y_A, z_B - z_A \rangle = \langle 1 - 1, 1 - 0, 0 - 1 \rangle = \langle 0; 1; -1 \rangle.$$

If $t = 0$ then $(x, y, z) = (x_A, y_A, z_A) = (1, 0, 1)$; if $t = 1$ then $(x, y, z) = (x_B, y_B, z_B) = (1, 1, 0)$.

Coordinate $x = 1$ is constant on the line $AB$.

Thus, a vector parametrization of $AB$ is $(x, y, z) = (1, t, 1 - t)$, $0 \leq t \leq 1$.

Line $BC$ has $(x_0, y_0, z_0) = (x_B, y_B, z_B) = (1, 1, 0)$,

$$\vec{l} = \overrightarrow{BC} = \langle x_C - x_B, y_C - y_B, z_C - z_B \rangle = \langle 0 - 1, 1 - 1, 1 - 0 \rangle = \langle -1; 0; 1 \rangle.$$

If $t = 0$ then $(x, y, z) = (x_B, y_B, z_B) = (1, 1, 0)$; if $t = 1$ then $(x, y, z) = (x_C, y_C, z_C) = (0, 1, 1)$.

Coordinate $y = 1$ is constant on the line $BC$.

Thus, a vector parametrization of $BC$ is $(x, y, z) = (1 - t, 1, t)$, $0 \leq t \leq 1$.

Line $CA$ has $(x_0, y_0, z_0) = (x_C, y_C, z_C) = (0, 1, 1)$,

$$\vec{l} = \overrightarrow{CA} = \langle x_A - x_C, y_A - y_C, z_A - z_C \rangle = \langle 1 - 0, 0 - 1, 1 - 1 \rangle = \langle 1; -1; 0 \rangle.$$

If $t = 0$ then $(x, y, z) = (x_C, y_C, z_C) = (0, 1, 1)$; if $t = 1$ then $(x, y, z) = (x_A, y_A, z_A) = (1, 0, 1)$.

Coordinate $z = 1$ is constant on the line $CA$.

Thus, a vector parametrization of $CA$ is $(x, y, z) = (t, 1 - t, 1)$, $0 \leq t \leq 1$. In figure colour of $AB$ is blue, colour of $BC$ is yellow, colour of $CA$ is green.