Let's work through part (a) using the points $(-2,-2)$, $(-3,0)$, and $(2,-4)$.

Part (a): Find the equation of the plane passing through the three points

1. Identify the three points: $P_1(-2, -2), \, P_2(-3, 0), \, P_3(2, -4)$

2. Form vectors from the points: We can form two vectors from the points: $\vec{v_1} = P_2 - P_1 = (-3 - (-2), 0 - (-2)) = (-1, 2)$ $\vec{v_2} = P_3 - P_1 = (2 - (-2), -4 - (-2)) = (4, -2)$

3. Find the cross product of the two vectors: The cross product of $\vec{v_1} = (-1, 2)$ and $\vec{v_2} = (4, -2)$ in 2D can be simplified as:

   \begin{vmatrix} -1 & 2 \\ 4 & -2 \end{vmatrix} = (-1)(-2) - (2)(4) = 2 - 8 = -6$$

4. Determine the equation of the line (in 2D): The general form of a line in 2D is: $Ax + By + C = 0$ Using the cross product result $(-6)$, we substitute the coefficients into the equation. The line passing through $P_1(-2, -2)$ can be written as: $-6x + By + C = 0$ To find $B$ and $C$, substitute the points back into the equation to get their values.

This is the general approach for finding the equation of the plane. Do you have any specific questions on this, or would you like more detailed steps?

Related Questions:

1. How do you find the equation of a plane given three points?
2. What is the formula for finding a cross product in 3D?
3. How can you determine if three points are collinear?
4. What does it mean if the cross product of two vectors is zero?
5. How can you find the normal vector to a plane?

Tip:

Always check for collinearity when working with three points to ensure they form a valid plane.