Get comprehensive solutions to your problems with IDNLearn.com. Get prompt and accurate answers to your questions from our experts who are always ready to help.
Sagot :
### 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.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.