Answer :
### Part A
To determine whether the vectors [tex]\(\overrightarrow{v_1} = \langle \sqrt{3}, 1 \rangle\)[/tex] and [tex]\(\overrightarrow{v_2} = \langle -\sqrt{3}, -1 \rangle\)[/tex] are parallel, we need to check if one vector is a scalar multiple of the other. In other words, if there exists a scalar [tex]\(k\)[/tex] such that:
[tex]\[ \overrightarrow{v_2} = k \cdot \overrightarrow{v_1} \][/tex]
Let's compare the components of the vectors [tex]\(\overrightarrow{v_1}\)[/tex] and [tex]\(\overrightarrow{v_2}\)[/tex].
The first component comparison:
[tex]\[ -\sqrt{3} = k \cdot \sqrt{3} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{-\sqrt{3}}{\sqrt{3}} = -1 \][/tex]
Now, let's verify using the second component:
[tex]\[ -1 = k \cdot 1 \Rightarrow -1 = -1 \][/tex]
Both components yield the same scalar [tex]\(k = -1\)[/tex]. When the ratio of the corresponding components of two vectors is the same, it implies that the vectors are parallel. Here the ratio is [tex]\([-1, -1]\)[/tex].
Therefore, the vectors [tex]\(\overrightarrow{v_1} = \langle \sqrt{3}, 1 \rangle\)[/tex] and [tex]\(\overrightarrow{v_2} = \langle -\sqrt{3}, -1 \rangle\)[/tex] are indeed parallel as both conditions are satisfied.
Conclusion: The vectors [tex]\(\overrightarrow{v_1}\)[/tex] and [tex]\(\overrightarrow{v_2}\)[/tex] are parallel.
### Part B
Similarly, to determine whether the vectors [tex]\(\overrightarrow{u_1} = \langle 2, 3 \rangle\)[/tex] and [tex]\(\overrightarrow{u_2} = \langle -3, -2 \rangle\)[/tex] are parallel, we should again check if one vector is a scalar multiple of the other. That is, if there exists a scalar [tex]\(k\)[/tex] such that:
[tex]\[ \overrightarrow{u_2} = k \cdot \overrightarrow{u_1} \][/tex]
Consider the first component:
[tex]\[ -3 = k \cdot 2 \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{-3}{2} = -1.5 \][/tex]
And now the second component:
[tex]\[ -2 = k \cdot 3 \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{-2}{3} \approx -0.6667 \][/tex]
Since the scalar [tex]\(k\)[/tex] is not consistently the same for both components (one component gives [tex]\(k = -1.5\)[/tex] and the other gives [tex]\(k \approx -0.6667\)[/tex]), the vectors are not scalar multiples of each other.
Conclusion: The vectors [tex]\(\overrightarrow{u_1} = \langle 2, 3 \rangle\)[/tex] and [tex]\(\overrightarrow{u_2} = \langle -3, -2 \rangle\)[/tex] are not parallel.
To determine whether the vectors [tex]\(\overrightarrow{v_1} = \langle \sqrt{3}, 1 \rangle\)[/tex] and [tex]\(\overrightarrow{v_2} = \langle -\sqrt{3}, -1 \rangle\)[/tex] are parallel, we need to check if one vector is a scalar multiple of the other. In other words, if there exists a scalar [tex]\(k\)[/tex] such that:
[tex]\[ \overrightarrow{v_2} = k \cdot \overrightarrow{v_1} \][/tex]
Let's compare the components of the vectors [tex]\(\overrightarrow{v_1}\)[/tex] and [tex]\(\overrightarrow{v_2}\)[/tex].
The first component comparison:
[tex]\[ -\sqrt{3} = k \cdot \sqrt{3} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{-\sqrt{3}}{\sqrt{3}} = -1 \][/tex]
Now, let's verify using the second component:
[tex]\[ -1 = k \cdot 1 \Rightarrow -1 = -1 \][/tex]
Both components yield the same scalar [tex]\(k = -1\)[/tex]. When the ratio of the corresponding components of two vectors is the same, it implies that the vectors are parallel. Here the ratio is [tex]\([-1, -1]\)[/tex].
Therefore, the vectors [tex]\(\overrightarrow{v_1} = \langle \sqrt{3}, 1 \rangle\)[/tex] and [tex]\(\overrightarrow{v_2} = \langle -\sqrt{3}, -1 \rangle\)[/tex] are indeed parallel as both conditions are satisfied.
Conclusion: The vectors [tex]\(\overrightarrow{v_1}\)[/tex] and [tex]\(\overrightarrow{v_2}\)[/tex] are parallel.
### Part B
Similarly, to determine whether the vectors [tex]\(\overrightarrow{u_1} = \langle 2, 3 \rangle\)[/tex] and [tex]\(\overrightarrow{u_2} = \langle -3, -2 \rangle\)[/tex] are parallel, we should again check if one vector is a scalar multiple of the other. That is, if there exists a scalar [tex]\(k\)[/tex] such that:
[tex]\[ \overrightarrow{u_2} = k \cdot \overrightarrow{u_1} \][/tex]
Consider the first component:
[tex]\[ -3 = k \cdot 2 \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{-3}{2} = -1.5 \][/tex]
And now the second component:
[tex]\[ -2 = k \cdot 3 \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{-2}{3} \approx -0.6667 \][/tex]
Since the scalar [tex]\(k\)[/tex] is not consistently the same for both components (one component gives [tex]\(k = -1.5\)[/tex] and the other gives [tex]\(k \approx -0.6667\)[/tex]), the vectors are not scalar multiples of each other.
Conclusion: The vectors [tex]\(\overrightarrow{u_1} = \langle 2, 3 \rangle\)[/tex] and [tex]\(\overrightarrow{u_2} = \langle -3, -2 \rangle\)[/tex] are not parallel.
