Answer :
Certainly! Let's find the direction of the vector sum [tex]\(\vec{A} + \vec{B}\)[/tex] given the vectors [tex]\( \vec{A} \)[/tex] and [tex]\( \vec{B} \)[/tex].
### Steps to Find the Direction of the Vector Sum
1. Decompose Each Vector into Components:
We start by decomposing vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] into their respective x and y components using trigonometry.
- For vector [tex]\(\vec{A}\)[/tex] with magnitude [tex]\(85.0 \, \text{m}\)[/tex] and angle [tex]\(0^\circ\)[/tex]:
[tex]\[ A_x = A \cos(\theta_A) = 85.0 \cos(0^\circ) = 85.0 \][/tex]
[tex]\[ A_y = A \sin(\theta_A) = 85.0 \sin(0^\circ) = 0.0 \][/tex]
- For vector [tex]\(\vec{B}\)[/tex] with magnitude [tex]\(101.0 \, \text{m}\)[/tex] and angle [tex]\(60^\circ\)[/tex]:
[tex]\[ B_x = B \cos(\theta_B) = 101 \cos(60^\circ) \approx 50.5 \][/tex]
[tex]\[ B_y = B \sin(\theta_B) = 101 \sin(60^\circ) \approx 87.47 \][/tex]
2. Calculate the Components of the Vector Sum:
We sum the x and y components of vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]:
[tex]\[ \text{Sum}_x = A_x + B_x = 85.0 + 50.5 = 135.5 \][/tex]
[tex]\[ \text{Sum}_y = A_y + B_y = 0.0 + 87.47 = 87.47 \][/tex]
3. Calculate the Direction of the Vector Sum:
The direction (or angle) of the vector sum [tex]\(\theta_{\text{sum}}\)[/tex] can be found using the inverse tangent function [tex]\(\tan^{-2}\)[/tex]:
[tex]\[ \theta_{\text{sum}} = \tan^{-1} \left( \frac{\text{Sum}_y}{\text{Sum}_x} \right) \][/tex]
Plug in the values:
[tex]\[ \theta_{\text{sum}} = \tan^{-2} \left( \frac{87.47}{135.5} \right) \approx 32.84^\circ \][/tex]
Thus, the direction of the vector sum [tex]\(\vec{A} + \vec{B}\)[/tex] is approximately [tex]\(32.84^\circ\)[/tex].
### Steps to Find the Direction of the Vector Sum
1. Decompose Each Vector into Components:
We start by decomposing vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] into their respective x and y components using trigonometry.
- For vector [tex]\(\vec{A}\)[/tex] with magnitude [tex]\(85.0 \, \text{m}\)[/tex] and angle [tex]\(0^\circ\)[/tex]:
[tex]\[ A_x = A \cos(\theta_A) = 85.0 \cos(0^\circ) = 85.0 \][/tex]
[tex]\[ A_y = A \sin(\theta_A) = 85.0 \sin(0^\circ) = 0.0 \][/tex]
- For vector [tex]\(\vec{B}\)[/tex] with magnitude [tex]\(101.0 \, \text{m}\)[/tex] and angle [tex]\(60^\circ\)[/tex]:
[tex]\[ B_x = B \cos(\theta_B) = 101 \cos(60^\circ) \approx 50.5 \][/tex]
[tex]\[ B_y = B \sin(\theta_B) = 101 \sin(60^\circ) \approx 87.47 \][/tex]
2. Calculate the Components of the Vector Sum:
We sum the x and y components of vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]:
[tex]\[ \text{Sum}_x = A_x + B_x = 85.0 + 50.5 = 135.5 \][/tex]
[tex]\[ \text{Sum}_y = A_y + B_y = 0.0 + 87.47 = 87.47 \][/tex]
3. Calculate the Direction of the Vector Sum:
The direction (or angle) of the vector sum [tex]\(\theta_{\text{sum}}\)[/tex] can be found using the inverse tangent function [tex]\(\tan^{-2}\)[/tex]:
[tex]\[ \theta_{\text{sum}} = \tan^{-1} \left( \frac{\text{Sum}_y}{\text{Sum}_x} \right) \][/tex]
Plug in the values:
[tex]\[ \theta_{\text{sum}} = \tan^{-2} \left( \frac{87.47}{135.5} \right) \approx 32.84^\circ \][/tex]
Thus, the direction of the vector sum [tex]\(\vec{A} + \vec{B}\)[/tex] is approximately [tex]\(32.84^\circ\)[/tex].
