Sure, let's find the vectors [tex]\(\vec{u}\)[/tex] and [tex]\(\vec{v}\)[/tex] given the vector equations:
1. [tex]\(\vec{u} + 2\vec{v} = 3\vec{i} - \vec{k}\)[/tex]
2. [tex]\(3\vec{u} - \vec{v} = \vec{i} + \vec{j} + \vec{k}\)[/tex]
First, let's express [tex]\(\vec{u}\)[/tex] and [tex]\(\vec{v}\)[/tex] in terms of their components:
[tex]\[
\vec{u} = u_1 \vec{i} + u_2 \vec{j} + u_3 \vec{k}
\][/tex]
[tex]\[
\vec{v} = v_1 \vec{i} + v_2 \vec{j} + v_3 \vec{k}
\][/tex]
Next, we will break down the vector equations into scalar components by comparing coefficients on both sides for each of the unit vectors [tex]\(\vec{i}\)[/tex], [tex]\(\vec{j}\)[/tex], and [tex]\(\vec{k}\)[/tex].
### Equation 1:
[tex]\[
\vec{u} + 2\vec{v} = 3\vec{i} - \vec{k}
\][/tex]
Comparing [tex]\(\vec{i}\)[/tex], [tex]\(\vec{j}\)[/tex], and [tex]\(\vec{k}\)[/tex] components:
[tex]\[
u_1 + 2v_1 = 3 \quad (1a)
\][/tex]
[tex]\[
u_2 + 2v_2 = 0 \quad (1b)
\][/tex]
[tex]\[
u_3 + 2v_3 = -1 \quad (1c)
\][/tex]
### Equation 2:
[tex]\[
3\vec{u} - \vec{v} = \vec{i} + \vec{j} + \vec{k}
\][/tex]
Comparing [tex]\(\vec{i}\)[/tex], [tex]\(\vec{j}\)[/tex], and [tex]\(\vec{k}\)[/tex] components:
[tex]\[
3u_1 - v_1 = 1 \quad (2a)
\][/tex]
[tex]\[
3u_2 - v_2 = 1 \quad (2b)
\][/tex]
[tex]\[
3u_3 - v_3 = 1 \quad (2c)
\][/tex]
We now have a system of linear equations:
For [tex]\(u_1\)[/tex] and [tex]\(v_1\)[/tex]:
[tex]\[
\begin{cases}
u_1 + 2v_1 = 3 \\
3u_1 - v_1 = 1
\end{cases}
\][/tex]
For [tex]\(u_2\)[/tex] and [tex]\(v_2\)[/tex]:
[tex]\[
\begin{cases}
u_2 + 2v_2 = 0 \\
3u_2 - v_2 = 1
\end{cases}
\][/tex]
For [tex]\(u_3\)[/tex] and [tex]\(v_3\)[/tex]:
[tex]\[
\begin{cases}
u_3 + 2v_3 = -1 \\
3u_3 - v_3 = 1
\end{cases}
\][/tex]
#### Solving these linear systems:
1. For [tex]\(u_1\)[/tex] and [tex]\(v_1\)[/tex]:
[tex]\[
\begin{cases}
u_1 + 2v_1 = 3 \\
3u_1 - v_1 = 1
\end{cases}
\][/tex]
Solving the above system, we get:
[tex]\(
u_1 = 0.85714286
\)[/tex]
[tex]\(
v_1 = 0.71428571
\)[/tex]
2. For [tex]\(u_2\)[/tex] and [tex]\(v_2\)[/tex]:
[tex]\[
\begin{cases}
u_2 + 2v_2 = 0 \\
3u_2 - v_2 = 1
\end{cases}
\][/tex]
Solving the above system, we get:
[tex]\(
u_2 = 0.42857143
\)[/tex]
[tex]\(
v_2 = -0.14285714
\)[/tex]
3. For [tex]\(u_3\)[/tex] and [tex]\(v_3\)[/tex]:
[tex]\[
\begin{cases}
u_3 + 2v_3 = -1 \\
3u_3 - v_3 = 1
\end{cases}
\][/tex]
Solving the above system, we get:
[tex]\(
u_3 = 0.28571429
\)[/tex]
[tex]\(
v_3 = -0.42857143
\)[/tex]
Thus, the vectors [tex]\(\vec{u}\)[/tex] and [tex]\(\vec{v}\)[/tex] are:
[tex]\[
\vec{u} = \left[0.85714286, 0.42857143, 0.28571429\right]
\][/tex]
[tex]\[
\vec{v} = \left[0.71428571, -0.14285714, -0.42857143\right]
\][/tex]
