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To determine both the vector equation and the Cartesian equation of the plane containing the given vectors and point, we go through the following steps:
### Step-by-Step Solution:
1. Define the Given Variables:
- Two vectors in the plane: [tex]\( \mathbf{v_1} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \)[/tex] and [tex]\( \mathbf{v_2} = \begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix} \)[/tex].
- A point on the plane: [tex]\( A(1, 2, 0) \)[/tex].
2. Find the Normal Vector to the Plane:
To find the normal vector to the plane, we compute the cross product of the given vectors [tex]\( \mathbf{v_1} \)[/tex] and [tex]\( \mathbf{v_2} \)[/tex].
[tex]\[ \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 3 & 0 & -1 \end{vmatrix} = \mathbf{i} \left( 1 \cdot (-1) - 1 \cdot 0 \right) - \mathbf{j} \left( 2 \cdot (-1) - 1 \cdot 3 \right) + \mathbf{k} \left( 2 \cdot 0 - 1 \cdot 3 \right) \][/tex]
[tex]\[ = \mathbf{i}(-1) - \mathbf{j}(-2 - 3) + \mathbf{k}(0 - 3) \][/tex]
[tex]\[ = -\mathbf{i} + 5\mathbf{j} - 3\mathbf{k} \][/tex]
Thus, the normal vector is [tex]\( \mathbf{n} = \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} \)[/tex].
3. Vector Equation of the Plane:
The vector equation of the plane can be represented with a normal vector [tex]\(\mathbf{n}\)[/tex] and a point [tex]\( A(1, 2, 0) \)[/tex] on the plane:
[tex]\[ \mathbf{n} \cdot \left(\mathbf{r} - \mathbf{r_0}\right) = 0 \][/tex]
where [tex]\(\mathbf{r}\)[/tex] is any point on the plane, and [tex]\(\mathbf{r_0} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\)[/tex] is the given point. Thus, the vector equation is:
[tex]\[ \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} \cdot \left( \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \right) = 0 \][/tex]
4. Simplify to Cartesian Equation:
Expand the dot product:
[tex]\[ \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} x - 1 \\ y - 2 \\ z - 0 \end{pmatrix} = 0 \][/tex]
[tex]\[ -1(x - 1) + 5(y - 2) - 3z = 0 \][/tex]
Simplify:
[tex]\[ -x + 1 + 5y - 10 - 3z = 0 \][/tex]
[tex]\[ -x + 5y - 3z - 9 = 0 \][/tex]
[tex]\[ -x + 5y - 3z = 9 \][/tex]
Hence, the Cartesian equation of the plane is:
[tex]\[ -1x + 5y + -3z + -9 = 0 \][/tex]
And in a cleaner format:
[tex]\[ -x + 5y - 3z - 9 = 0 \][/tex]
The final results are:
- Vector equation of the plane:
[tex]\[ \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} \cdot \left( \mathbf{r} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \right) = 0 \][/tex]
- Cartesian equation of the plane:
[tex]\[ -x + 5y - 3z - 9 = 0 \][/tex]
### Step-by-Step Solution:
1. Define the Given Variables:
- Two vectors in the plane: [tex]\( \mathbf{v_1} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \)[/tex] and [tex]\( \mathbf{v_2} = \begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix} \)[/tex].
- A point on the plane: [tex]\( A(1, 2, 0) \)[/tex].
2. Find the Normal Vector to the Plane:
To find the normal vector to the plane, we compute the cross product of the given vectors [tex]\( \mathbf{v_1} \)[/tex] and [tex]\( \mathbf{v_2} \)[/tex].
[tex]\[ \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 3 & 0 & -1 \end{vmatrix} = \mathbf{i} \left( 1 \cdot (-1) - 1 \cdot 0 \right) - \mathbf{j} \left( 2 \cdot (-1) - 1 \cdot 3 \right) + \mathbf{k} \left( 2 \cdot 0 - 1 \cdot 3 \right) \][/tex]
[tex]\[ = \mathbf{i}(-1) - \mathbf{j}(-2 - 3) + \mathbf{k}(0 - 3) \][/tex]
[tex]\[ = -\mathbf{i} + 5\mathbf{j} - 3\mathbf{k} \][/tex]
Thus, the normal vector is [tex]\( \mathbf{n} = \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} \)[/tex].
3. Vector Equation of the Plane:
The vector equation of the plane can be represented with a normal vector [tex]\(\mathbf{n}\)[/tex] and a point [tex]\( A(1, 2, 0) \)[/tex] on the plane:
[tex]\[ \mathbf{n} \cdot \left(\mathbf{r} - \mathbf{r_0}\right) = 0 \][/tex]
where [tex]\(\mathbf{r}\)[/tex] is any point on the plane, and [tex]\(\mathbf{r_0} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\)[/tex] is the given point. Thus, the vector equation is:
[tex]\[ \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} \cdot \left( \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \right) = 0 \][/tex]
4. Simplify to Cartesian Equation:
Expand the dot product:
[tex]\[ \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} x - 1 \\ y - 2 \\ z - 0 \end{pmatrix} = 0 \][/tex]
[tex]\[ -1(x - 1) + 5(y - 2) - 3z = 0 \][/tex]
Simplify:
[tex]\[ -x + 1 + 5y - 10 - 3z = 0 \][/tex]
[tex]\[ -x + 5y - 3z - 9 = 0 \][/tex]
[tex]\[ -x + 5y - 3z = 9 \][/tex]
Hence, the Cartesian equation of the plane is:
[tex]\[ -1x + 5y + -3z + -9 = 0 \][/tex]
And in a cleaner format:
[tex]\[ -x + 5y - 3z - 9 = 0 \][/tex]
The final results are:
- Vector equation of the plane:
[tex]\[ \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} \cdot \left( \mathbf{r} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \right) = 0 \][/tex]
- Cartesian equation of the plane:
[tex]\[ -x + 5y - 3z - 9 = 0 \][/tex]
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