Answer :
Sure, let's solve the problem step by step for adding and subtracting vectors, including finding their magnitudes and directions.
### Part (a): Finding [tex]\(|A - B|\)[/tex] and its direction
1. Representing Vectors in Cartesian Coordinates:
- Vector A:
- Magnitude [tex]\( A = 26.0 \)[/tex] m
- Angle [tex]\( \theta_A = 30^\circ \)[/tex] above the positive x-axis
[tex]\[ A_x = A \cos(\theta_A) = 26 \cos(30^\circ) \][/tex]
[tex]\[ A_y = A \sin(\theta_A) = 26 \sin(30^\circ) \][/tex]
Since [tex]\( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)[/tex] and [tex]\( \sin(30^\circ) = \frac{1}{2} \)[/tex]:
[tex]\[ A_x = 26 \times \frac{\sqrt{3}}{2} = 13\sqrt{3} \][/tex]
[tex]\[ A_y = 26 \times \frac{1}{2} = 13 \][/tex]
- Vector B:
- Magnitude [tex]\( B = 11.0 \)[/tex] m
- Orientation: [tex]\( 60^\circ \)[/tex] to the left of the y-axis means [tex]\( 60^\circ + 90^\circ = 150^\circ \)[/tex] from the positive x-axis in the counterclockwise direction.
[tex]\[ B_x = B \cos(\theta_B) = 11 \cos(150^\circ) \][/tex]
[tex]\[ B_y = B \sin(\theta_B) = 11 \sin(150^\circ) \][/tex]
Using the trigonometric values, [tex]\(\cos(150^\circ) = -\frac{\sqrt{3}}{2}\)[/tex] and [tex]\(\sin(150^\circ) = \frac{1}{2}\)[/tex]:
[tex]\[ B_x = 11 \times -\frac{\sqrt{3}}{2} = -5.5\sqrt{3} \][/tex]
[tex]\[ B_y = 11 \times \frac{1}{2} = 5.5 \][/tex]
2. Finding Components of [tex]\( A - B \)[/tex]:
- Subtract the corresponding components of B from A.
[tex]\[ (A - B)_x = A_x - B_x = 13\sqrt{3} - (-5.5\sqrt{3}) = 13\sqrt{3} + 5.5\sqrt{3} = 18.5\sqrt{3} \][/tex]
[tex]\[ (A - B)_y = A_y - B_y = 13 - 5.5 = 7.5 \][/tex]
3. Magnitude of [tex]\( A - B \)[/tex]:
[tex]\[ |A - B| = \sqrt{(A - B)_x^2 + (A - B)_y^2} = \sqrt{(18.5\sqrt{3})^2 + (7.5)^2} \][/tex]
Resulting in approximately [tex]\( 32.91 \)[/tex] m (from our preliminary computations).
4. Direction of [tex]\( A - B \)[/tex]:
[tex]\[ \theta_{A-B} = \tan^{-1} \left( \frac{(A - B)_y}{(A - B)_x} \right) = \tan^{-1} \left( \frac{7.5}{18.5\sqrt{3}} \right) \][/tex]
Resulting in approximately [tex]\( 13.17^\circ \)[/tex].
### Part (b): Finding [tex]\(|2A + B|\)[/tex] and its direction
1. Components of [tex]\( 2A \)[/tex]:
- Multiply each component of A by 2.
[tex]\[ (2A)_x = 2 \times A_x = 2 \times 13\sqrt{3} = 26\sqrt{3} \][/tex]
[tex]\[ (2A)_y = 2 \times A_y = 2 \times 13 = 26 \][/tex]
2. Finding Components of [tex]\( 2A + B \)[/tex]:
- Add the corresponding components of B to [tex]\(2A\)[/tex].
[tex]\[ (2A + B)_x = (2A)_x + B_x = 26\sqrt{3} + (-5.5\sqrt{3}) = 20.5\sqrt{3} \][/tex]
[tex]\[ (2A + B)_y = (2A)_y + B_y = 26 + 5.5 = 31.5 \][/tex]
3. Magnitude of [tex]\( 2A + B \)[/tex]:
[tex]\[ |2A + B| = \sqrt{((2A + B)_x)^2 + ((2A + B)_y)^2} = \sqrt{(20.5\sqrt{3})^2 + 31.5^2} \][/tex]
Resulting in approximately [tex]\( 47.47 \)[/tex] m.
4. Direction of [tex]\( 2A + B \)[/tex]:
[tex]\[ \theta_{2A+B} = \tan^{-1} \left( \frac{(2A + B)_y}{(2A + B)_x} \right) = \tan^{-1} \left( \frac{31.5}{20.5\sqrt{3}} \right) \][/tex]
Resulting in approximately [tex]\( 41.58^\circ \)[/tex].
To summarize:
- [tex]\(|A - B| \approx 32.91\)[/tex] m and direction [tex]\(\approx 13.17^\circ\)[/tex]
- [tex]\(|2A + B| \approx 47.47\)[/tex] m and direction [tex]\(\approx 41.58^\circ\)[/tex]
These results give the magnitude and direction of the vectors as required.
### Part (a): Finding [tex]\(|A - B|\)[/tex] and its direction
1. Representing Vectors in Cartesian Coordinates:
- Vector A:
- Magnitude [tex]\( A = 26.0 \)[/tex] m
- Angle [tex]\( \theta_A = 30^\circ \)[/tex] above the positive x-axis
[tex]\[ A_x = A \cos(\theta_A) = 26 \cos(30^\circ) \][/tex]
[tex]\[ A_y = A \sin(\theta_A) = 26 \sin(30^\circ) \][/tex]
Since [tex]\( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)[/tex] and [tex]\( \sin(30^\circ) = \frac{1}{2} \)[/tex]:
[tex]\[ A_x = 26 \times \frac{\sqrt{3}}{2} = 13\sqrt{3} \][/tex]
[tex]\[ A_y = 26 \times \frac{1}{2} = 13 \][/tex]
- Vector B:
- Magnitude [tex]\( B = 11.0 \)[/tex] m
- Orientation: [tex]\( 60^\circ \)[/tex] to the left of the y-axis means [tex]\( 60^\circ + 90^\circ = 150^\circ \)[/tex] from the positive x-axis in the counterclockwise direction.
[tex]\[ B_x = B \cos(\theta_B) = 11 \cos(150^\circ) \][/tex]
[tex]\[ B_y = B \sin(\theta_B) = 11 \sin(150^\circ) \][/tex]
Using the trigonometric values, [tex]\(\cos(150^\circ) = -\frac{\sqrt{3}}{2}\)[/tex] and [tex]\(\sin(150^\circ) = \frac{1}{2}\)[/tex]:
[tex]\[ B_x = 11 \times -\frac{\sqrt{3}}{2} = -5.5\sqrt{3} \][/tex]
[tex]\[ B_y = 11 \times \frac{1}{2} = 5.5 \][/tex]
2. Finding Components of [tex]\( A - B \)[/tex]:
- Subtract the corresponding components of B from A.
[tex]\[ (A - B)_x = A_x - B_x = 13\sqrt{3} - (-5.5\sqrt{3}) = 13\sqrt{3} + 5.5\sqrt{3} = 18.5\sqrt{3} \][/tex]
[tex]\[ (A - B)_y = A_y - B_y = 13 - 5.5 = 7.5 \][/tex]
3. Magnitude of [tex]\( A - B \)[/tex]:
[tex]\[ |A - B| = \sqrt{(A - B)_x^2 + (A - B)_y^2} = \sqrt{(18.5\sqrt{3})^2 + (7.5)^2} \][/tex]
Resulting in approximately [tex]\( 32.91 \)[/tex] m (from our preliminary computations).
4. Direction of [tex]\( A - B \)[/tex]:
[tex]\[ \theta_{A-B} = \tan^{-1} \left( \frac{(A - B)_y}{(A - B)_x} \right) = \tan^{-1} \left( \frac{7.5}{18.5\sqrt{3}} \right) \][/tex]
Resulting in approximately [tex]\( 13.17^\circ \)[/tex].
### Part (b): Finding [tex]\(|2A + B|\)[/tex] and its direction
1. Components of [tex]\( 2A \)[/tex]:
- Multiply each component of A by 2.
[tex]\[ (2A)_x = 2 \times A_x = 2 \times 13\sqrt{3} = 26\sqrt{3} \][/tex]
[tex]\[ (2A)_y = 2 \times A_y = 2 \times 13 = 26 \][/tex]
2. Finding Components of [tex]\( 2A + B \)[/tex]:
- Add the corresponding components of B to [tex]\(2A\)[/tex].
[tex]\[ (2A + B)_x = (2A)_x + B_x = 26\sqrt{3} + (-5.5\sqrt{3}) = 20.5\sqrt{3} \][/tex]
[tex]\[ (2A + B)_y = (2A)_y + B_y = 26 + 5.5 = 31.5 \][/tex]
3. Magnitude of [tex]\( 2A + B \)[/tex]:
[tex]\[ |2A + B| = \sqrt{((2A + B)_x)^2 + ((2A + B)_y)^2} = \sqrt{(20.5\sqrt{3})^2 + 31.5^2} \][/tex]
Resulting in approximately [tex]\( 47.47 \)[/tex] m.
4. Direction of [tex]\( 2A + B \)[/tex]:
[tex]\[ \theta_{2A+B} = \tan^{-1} \left( \frac{(2A + B)_y}{(2A + B)_x} \right) = \tan^{-1} \left( \frac{31.5}{20.5\sqrt{3}} \right) \][/tex]
Resulting in approximately [tex]\( 41.58^\circ \)[/tex].
To summarize:
- [tex]\(|A - B| \approx 32.91\)[/tex] m and direction [tex]\(\approx 13.17^\circ\)[/tex]
- [tex]\(|2A + B| \approx 47.47\)[/tex] m and direction [tex]\(\approx 41.58^\circ\)[/tex]
These results give the magnitude and direction of the vectors as required.
