Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
