Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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 time. Please come back anytime for the latest information and answers to your questions. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
