To solve the given problem, we need to determine the vector, parametric, and general (Cartesian) equations of the plane defined by the points \( A(3,4,-1) \), \( B(2,-3,4) \), and \( C(-1,-1,5) \).
- Find two vectors that lie on the plane using the given points.
- Use one of the points and the two vectors to write the vector equation of the plane.
- Use the vector equation to express the coordinates of any point on the plane in terms of two parameters.
- Find the normal vector to the plane by taking the cross product of the two vectors found in part (a).
- Use the normal vector and one of the points to write the general equation of the plane.
La ecuación vectorial de un plano que pasa por los puntos \( A(3, 4, -1) \), \( B(2, -3, 4) \) y \( C(-1, -1, 5) \) se puede expresar como:
\[
\mathbf{r} = \begin{bmatrix} 3 \\ 4 \\ -1 \end{bmatrix} + s \cdot \begin{bmatrix} -1 \\ -7 \\ 5 \end{bmatrix} + t \cdot \begin{bmatrix} -4 \\ -5 \\ 6 \end{bmatrix}
\]
La ecuación paramétrica del plano se puede descomponer en sus componentes \( x \), \( y \) y \( z \):
\[
\begin{align_}
x &= 3 - s - 4t \\
y &= 4 - 7s - 5t \\
z &= -1 + 5s + 6t
\end{align_}
\]
La ecuación general del plano se obtiene utilizando el vector normal \( \mathbf{n} = \begin{bmatrix} -17 \\ -14 \\ -23 \end{bmatrix} \) y un punto en el plano. La forma general es:
\[
-17x - 14y - 23z + 84 = 0
\]
\[
\boxed{
\begin{align_}
\text{Ecuación Vectorial:} & \quad \mathbf{r} = \begin{bmatrix} 3 \\ 4 \\ -1 \end{bmatrix} + s \cdot \begin{bmatrix} -1 \\ -7 \\ 5 \end{bmatrix} + t \cdot \begin{bmatrix} -4 \\ -5 \\ 6 \end{bmatrix} \\
\text{Ecuación Paramétrica:} & \quad x = 3 - s - 4t, \quad y = 4 - 7s - 5t, \quad z = -1 + 5s + 6t \\
\text{Ecuación General:} & \quad -17x - 14y - 23z + 84 = 0
\end{align_}
}
\]
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