Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To draw the vector [tex]\( 2\mathbf{u} - \frac{1}{3}\mathbf{v} \)[/tex], let's go through the steps methodically. We are given two vectors [tex]\(\mathbf{u}\)[/tex] and [tex]\(\mathbf{v}\)[/tex]:
1. Identify the vectors:
[tex]\[ \mathbf{u} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad \mathbf{v} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \][/tex]
2. Multiply vectors by their respective scalars:
- Multiply [tex]\(\mathbf{u}\)[/tex] by 2:
[tex]\[ 2\mathbf{u} = 2 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} \][/tex]
- Multiply [tex]\(\mathbf{v}\)[/tex] by [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[ \frac{1}{3}\mathbf{v} = \frac{1}{3} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{1}{3} \end{pmatrix} \][/tex]
3. Subtract [tex]\(\frac{1}{3}\mathbf{v}\)[/tex] from 2[tex]\(\mathbf{u}\)[/tex]:
[tex]\[ 2\mathbf{u} - \frac{1}{3}\mathbf{v} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ \frac{1}{3} \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ -\frac{1}{3} \end{pmatrix} = \begin{pmatrix} 2 \\ -\frac{1}{3} \end{pmatrix} \][/tex]
4. Resultant vector:
[tex]\[ 2\mathbf{u} - \frac{1}{3}\mathbf{v} = \begin{pmatrix} 2 \\ -\frac{1}{3} \end{pmatrix} \][/tex]
Now, to visualize this on a coordinate plane:
- Plot the initial vectors [tex]\(\mathbf{u}\)[/tex] and [tex]\(\mathbf{v}\)[/tex]:
- [tex]\(\mathbf{u} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\)[/tex] starts at the origin (0,0) and extends to the point (1,0).
- [tex]\(\mathbf{v} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\)[/tex] starts at the origin (0,0) and extends to the point (0,1).
- Vector [tex]\(2\mathbf{u}\)[/tex]:
- Doubling [tex]\(\mathbf{u}\)[/tex] results in the vector from (0,0) to (2,0).
- Vector [tex]\(\frac{1}{3}\mathbf{v}\)[/tex]:
- Scaling [tex]\(\mathbf{v}\)[/tex] by [tex]\(\frac{1}{3}\)[/tex] results in the vector from (0,0) to (0, [tex]\(\frac{1}{3}\)[/tex]).
- Apply the subtraction:
- The final vector [tex]\(2\mathbf{u} - \frac{1}{3}\mathbf{v}\)[/tex] moves from (0,0) to (2, [tex]\(-\frac{1}{3}\)[/tex]).
To draw it:
1. Start at the origin (0,0).
2. Move 2 units to the right to reach the point (2,0).
3. Move [tex]\(\frac{1}{3}\)[/tex] units down (since the y-component is negative) to reach the point (2, [tex]\(-\frac{1}{3}\)[/tex]).
The vector [tex]\(2\mathbf{u} - \frac{1}{3}\mathbf{v}\)[/tex] can be represented as an arrow starting from the origin and ending at the coordinates (2, [tex]\(-\frac{1}{3}\)[/tex]).
1. Identify the vectors:
[tex]\[ \mathbf{u} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad \mathbf{v} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \][/tex]
2. Multiply vectors by their respective scalars:
- Multiply [tex]\(\mathbf{u}\)[/tex] by 2:
[tex]\[ 2\mathbf{u} = 2 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} \][/tex]
- Multiply [tex]\(\mathbf{v}\)[/tex] by [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[ \frac{1}{3}\mathbf{v} = \frac{1}{3} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{1}{3} \end{pmatrix} \][/tex]
3. Subtract [tex]\(\frac{1}{3}\mathbf{v}\)[/tex] from 2[tex]\(\mathbf{u}\)[/tex]:
[tex]\[ 2\mathbf{u} - \frac{1}{3}\mathbf{v} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ \frac{1}{3} \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ -\frac{1}{3} \end{pmatrix} = \begin{pmatrix} 2 \\ -\frac{1}{3} \end{pmatrix} \][/tex]
4. Resultant vector:
[tex]\[ 2\mathbf{u} - \frac{1}{3}\mathbf{v} = \begin{pmatrix} 2 \\ -\frac{1}{3} \end{pmatrix} \][/tex]
Now, to visualize this on a coordinate plane:
- Plot the initial vectors [tex]\(\mathbf{u}\)[/tex] and [tex]\(\mathbf{v}\)[/tex]:
- [tex]\(\mathbf{u} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\)[/tex] starts at the origin (0,0) and extends to the point (1,0).
- [tex]\(\mathbf{v} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\)[/tex] starts at the origin (0,0) and extends to the point (0,1).
- Vector [tex]\(2\mathbf{u}\)[/tex]:
- Doubling [tex]\(\mathbf{u}\)[/tex] results in the vector from (0,0) to (2,0).
- Vector [tex]\(\frac{1}{3}\mathbf{v}\)[/tex]:
- Scaling [tex]\(\mathbf{v}\)[/tex] by [tex]\(\frac{1}{3}\)[/tex] results in the vector from (0,0) to (0, [tex]\(\frac{1}{3}\)[/tex]).
- Apply the subtraction:
- The final vector [tex]\(2\mathbf{u} - \frac{1}{3}\mathbf{v}\)[/tex] moves from (0,0) to (2, [tex]\(-\frac{1}{3}\)[/tex]).
To draw it:
1. Start at the origin (0,0).
2. Move 2 units to the right to reach the point (2,0).
3. Move [tex]\(\frac{1}{3}\)[/tex] units down (since the y-component is negative) to reach the point (2, [tex]\(-\frac{1}{3}\)[/tex]).
The vector [tex]\(2\mathbf{u} - \frac{1}{3}\mathbf{v}\)[/tex] can be represented as an arrow starting from the origin and ending at the coordinates (2, [tex]\(-\frac{1}{3}\)[/tex]).
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.