To determine which of the given vectors has a magnitude of [tex]\(\sqrt{65}\)[/tex] and a direction of [tex]\(\theta = 240.255^\circ\)[/tex], we need to calculate the magnitude and direction for each vector and compare them with the given requirements.
### Step 1: Calculate the Magnitude
The magnitude of a vector [tex]\(\mathbf{u} = \langle a, b \rangle\)[/tex] is given by:
[tex]\[
\|\mathbf{u}\| = \sqrt{a^2 + b^2}
\][/tex]
### Step 2: Calculate the Direction
The direction (or angle) [tex]\(\theta\)[/tex] of a vector [tex]\(\mathbf{u} = \langle a, b \rangle\)[/tex] is given by:
[tex]\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\][/tex]
Note: Make sure to account for the quadrant in which the vector lies, as [tex]\(\tan^{-1}\)[/tex] will only give angles within the range [tex]\(-90^\circ\)[/tex] to [tex]\(90^\circ\)[/tex].
### Vectors Analysis
Let's analyze each vector one by one.
#### Vector [tex]\( \mathbf{u} = \langle -7, -4 \rangle \)[/tex]
1. Magnitude:
[tex]\[
\|\mathbf{u}\| = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}
\][/tex]
2. Direction:
[tex]\[
\theta = \tan^{-1}\left(\frac{-4}{-7}\right) = \tan^{-1}\left(\frac{4}{7}\right)
\][/tex]
Since both components are negative, the vector is in the third quadrant. By adding [tex]\(180^\circ\)[/tex]:
[tex]\[
\theta \approx \tan^{-1}(0.5714) + 180^\circ \approx 29.745^\circ + 180^\circ \approx 209.745^\circ
\][/tex]
#### Vector [tex]\( \mathbf{u} = \langle -6, -4 \rangle \)[/tex]
1. Magnitude:
[tex]\[
\|\mathbf{u}\| = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}
\][/tex]
This magnitude does not match [tex]\(\sqrt{65}\)[/tex].
#### Vector [tex]\( \mathbf{u} = \langle -4, -6 \rangle \)[/tex]
1. Magnitude:
[tex]\[
\|\mathbf{u}\| = \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52}
\][/tex]
This magnitude does not match [tex]\(\sqrt{65}\)[/tex].
#### Vector [tex]\( \mathbf{u} = \langle -4, -7 \rangle \)[/tex]
1. Magnitude:
[tex]\[
\|\mathbf{u}\| = \sqrt{(-4)^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65}
\][/tex]
2. Direction:
[tex]\[
\theta = \tan^{-1}\left(\frac{-7}{-4}\right) = \tan^{-1}\left(\frac{7}{4}\right)
\][/tex]
Since both components are negative, the vector lies in the third quadrant. By adding [tex]\(180^\circ\)[/tex]:
[tex]\[
\theta \approx \tan^{-1}(1.75) + 180^\circ \approx 60.255^\circ + 180^\circ \approx 240.255^\circ
\][/tex]
### Conclusion
The vector [tex]\(\mathbf{u} = \langle -4, -7 \rangle\)[/tex] has a magnitude of [tex]\(\sqrt{65}\)[/tex] and a direction of [tex]\(\theta = 240.255^\circ\)[/tex], meeting the given criteria. None of the other vectors match both criteria.
