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We are tasked with converting the given plane equation \( -5x - 5y - 4z = 6 \) into vector parametric form.

The general form for a plane is:

\[
Ax + By + Cz = D
\]

The parametric form of a plane can be written as:

\[
\mathbf{r} = \mathbf{r}_0 + s\mathbf{d} + t\mathbf{e}
\]

Where:
- \( \mathbf{r}_0 \) is a point on the plane.
- \( \mathbf{d} \) and \( \mathbf{e} \) are two direction vectors parallel to the plane.
- \( s \) and \( t \) are parameters.

### Step 1: Find a Point on the Plane

We can choose values for two variables and solve for the third. Let's choose \( x = 0 \) and \( y = 0 \) to find a point \( \mathbf{r}_0 \).

Substitute \( x = 0 \) and \( y = 0 \) into the equation:

\[
-4z = 6 \quad \Rightarrow \quad z = -\frac{3}{2}
\]

Thus, one point on the plane is \( \mathbf{r}_0 = (0, 0, -\frac{3}{2}) \).

### Step 2: Find Two Direction Vectors

Now, we need two independent direction vectors that lie on the plane. To do this, we can assign values to two variables and solve for the third to find two more points.

1. Set \( x = 1 \) and \( y = 0 \):

\[
-5(1) - 4z = 6 \quad \Rightarrow \quad -5 - 4z = 6 \quad \Rightarrow \quad -4z = 11 \quad \Rightarrow \quad z = -\frac{11}{4}
\]

This gives the point \( (1, 0, -\frac{11}{4}) \).

2. Set \( x = 0 \) and \( y = 1 \):

\[
-5(1) - 4z = 6 \quad \Rightarrow \quad -5 - 4z = 6 \quad \Rightarrow \quad z = =- insert

Write the following plane in vector parametric form: -5x - 5y - 4z = 6

Learn how to convert the plane equation -5x - 5y - 4z = 6 into vector parametric form with detailed steps. This solution explains the process of finding a point on the plane and two direction vectors parallel to the plane. Suitable for university-level linear algebra students.

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