### Question 3

Two vectors are said to be parallel if they point in the same direction or if they point in opposite directions.

Part A
Are the vectors [tex]\overrightarrow{v_1} = \langle \sqrt{3}, 1 \rangle[/tex] and [tex]\overrightarrow{v_2} = \langle -\sqrt{3}, -1 \rangle[/tex] parallel? Show your work and explain.

Part B
Are the vectors [tex]\overrightarrow{u_1} = \langle 2, 3 \rangle[/tex] and [tex]\overrightarrow{u_2} = \langle -3, -2 \rangle[/tex] parallel? Show your work and explain.



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.