1. **Problem:** Given the parametric form of a plane $E$, find its coordinate form (Cartesian equation). 2. **Formula and rules:** The parametric form of a plane is given by: $$\vec{x} = \vec{p} + r\vec{u} + s\vec{v}$$ where $\vec{p}$ is a point on the plane, and $\vec{u}, \vec{v}$ are direction vectors. To find the coordinate form, we need a normal vector $\vec{n}$ perpendicular to both $\vec{u}$ and $\vec{v}$: $$\vec{n} = \vec{u} \times \vec{v}$$ The coordinate form is: $$n_1 x_1 + n_2 x_2 + n_3 x_3 = d$$ where $d = \vec{n} \cdot \vec{p}$. 3. **Step-by-step solution for (a):** - Given: $$\vec{p} = \begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}, \quad \vec{u} = \begin{pmatrix}4 \\ 3 \\ 0\end{pmatrix}, \quad \vec{v} = \begin{pmatrix}1 \\ 0 \\ 3\end{pmatrix}$$ - Compute the normal vector: $$\vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & 0 \\ 1 & 0 & 3 \end{vmatrix} = \begin{pmatrix}3 \cdot 3 - 0 \cdot 0 \\ 0 \cdot 1 - 4 \cdot 3 \\ 4 \cdot 0 - 3 \cdot 1\end{pmatrix} = \begin{pmatrix}9 \\ -12 \\ -3\end{pmatrix}$$ - Simplify by dividing by 3: $$\vec{n} = \begin{pmatrix}\cancel{9}/3 \\ \cancel{-12}/3 \\ \cancel{-3}/3\end{pmatrix} = \begin{pmatrix}3 \\ -4 \\ -1\end{pmatrix}$$ - Calculate $d$: $$d = \vec{n} \cdot \vec{p} = 3 \cdot 1 + (-4) \cdot (-1) + (-1) \cdot (-1) = 3 + 4 + 1 = 8$$ - Final coordinate form: $$3x_1 - 4x_2 - x_3 = 8$$ **Answer:** The coordinate form of plane $E$ is $$3x_1 - 4x_2 - x_3 = 8$$.