To parameterize the plane through the point $(3, 2, 2)$ with the normal vector $\langle -2, 4, -4 \rangle$, we need to find two independent direction vectors that lie in the plane and then express the general form of a point on the plane.

Step 1: Direction vectors

The normal vector $\langle -2, 4, -4 \rangle$ is orthogonal to every vector in the plane. To find two independent direction vectors, we can choose vectors perpendicular to $\langle -2, 4, -4 \rangle$.

Vector 1 ($\mathbf{v_1}$):

Let’s choose a vector $\mathbf{v_1} = \langle 1, 0, 0 \rangle$. Compute the dot product with $\langle -2, 4, -4 \rangle$: $\mathbf{v_1} \cdot \langle -2, 4, -4 \rangle = -2(1) + 4(0) - 4(0) = -2 \neq 0.$ This choice does not work; we need another vector.

To parameterize the plane through the point $(3, 2, 2)$ with the normal vector $\langle -2, 4, -4 \rangle$, we proceed as follows:

Step 1: Find Two Direction Vectors in the Plane

The plane's normal vector, $\langle -2, 4, -4 \rangle$, is perpendicular to all vectors lying in the plane. To find two such independent vectors, we choose vectors that are not multiples of the normal vector and satisfy the orthogonality condition:

Vector 1 ($\mathbf{v_1}$):

Choose $\mathbf{v_1} = \langle 2, 1, 0 \rangle$. Compute the dot product with $\langle -2, 4, -4 \rangle$: $\mathbf{v_1} \cdot \langle -2, 4, -4 \rangle = (-2)(2) + (4)(1) + (-4)(0) = -4 + 4 + 0 = 0.$ Thus, $\mathbf{v_1} = \langle 2, 1, 0 \rangle$ lies in the plane.

Vector 2 ($\mathbf{v_2}$):

Choose $\mathbf{v_2} = \langle 0, 1, 1 \rangle$. Compute the dot product with $\langle -2, 4, -4 \rangle$: $\mathbf{v_2} \cdot \langle -2, 4, -4 \rangle = (-2)(0) + (4)(1) + (-4)(1) = 0 + 4 - 4 = 0.$ Thus, $\mathbf{v_2} = \langle 0, 1, 1 \rangle$ also lies in the plane.

Step 2: Write the Parametric Equation

Using the point $(3, 2, 2)$ and the two direction vectors $\mathbf{v_1}$ and $\mathbf{v_2}$, the parameterization of the plane is: $\mathbf{r}(s, t) = \langle 3, 2, 2 \rangle + s \langle 2, 1, 0 \rangle + t \langle 0, 1, 1 \rangle.$

Simplify: $\mathbf{r}(s, t) = \langle 3 + 2s, 2 + s + t, 2 + t \rangle.$

Final Answer:

$\mathbf{r}(s, t) = \langle 3 + 2s, 2 + s + t, 2 + t \rangle.$

Let me know if you'd like a deeper explanation or additional details!

Follow-up Questions:

1. What is the significance of choosing the two direction vectors?
2. How can you verify that a parameterized point lies in the plane?
3. Could the parameterization change if we chose different direction vectors?
4. How is the normal vector related to the plane's equation?
5. How would you find the equation of the plane in Cartesian form?

Tip:

Always double-check the orthogonality of your direction vectors with the normal vector to ensure they lie in the plane!