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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]).
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