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