Vector A has a magnitude of 20m and makes an angle of 30 degrees above the positive x axis. Vector B--> has a magnitude of 15m and is oriented 60 degree to the left of the y axis.
Find the magnitude and direction of A--> - B-->. The magnitude and direction of 2A--> + b--> and the magnitude of -A--> + 3B-->
To find the magnitude and direction of A--> - B-->, we need to subtract the components of B--> from A-->.
Step 1: Convert the given information into component form for vectors A--> and B-->.
Vector A-->:
Magnitude (A): 20m
Angle (θ): 30 degrees above the positive x-axis
Convert to component form: A-->(Ax, Ay) = (20cos(30), 20sin(30))
Vector B-->:
Magnitude (B): 15m
Angle (θ): 60 degrees to the left of the y-axis
Convert to component form: B-->(Bx, By) = (-15sin(60), -15cos(60))
Step 2: Subtract the respective components of B--> from A--> to get the resultant vector A--> - B-->.
(A - B)-->(Ax - Bx, Ay - By) = (20cos(30) - (-15sin(60)), 20sin(30) - (-15cos(60)))
Step 3: Simplify the components obtained in Step 2.
(A - B)-->(20√3 - (-15/2), 10 - (-15√3/2))
(A - B)-->(20√3 + 15/2, 10 + 15√3/2)
Step 4: Find the magnitude of A--> - B--> using the Pythagorean theorem.
Magnitude (|A - B|) = √((20√3 + 15/2)^2 + (10 + 15√3/2)^2)
To find the magnitude and direction of 2A--> + B-->, we need to add twice the components of A--> to the components of B-->.
Step 5: Calculate 2A--> + B-->.
(2A + B)-->(2Ax + Bx, 2Ay + By) = (2*(20cos(30)) + (-15sin(60)), 2*(20sin(30)) + (-15cos(60)))
Step 6: Simplify the components obtained in Step 5.
(2A + B)-->(40√3 - 15/2, 20 - 15√3/2)
Step 7: Find the magnitude of 2A--> + B--> using the Pythagorean theorem.
Magnitude (|2A + B|) = √((40√3 - 15/2)^2 + (20 - 15√3/2)^2)
To find the magnitude of -A--> + 3B-->, we need to subtract the components of A--> from three times the components of B-->.
Step 8: Calculate -A--> + 3B-->.
(-A + 3B)-->(-Ax + 3Bx, -Ay + 3By) = (-1*(20cos(30)) + 3*(-15sin(60)), -1*(20sin(30)) + 3*(-15cos(60)))
Step 9: Simplify the components obtained in Step 8.
(-A + 3B)-->(-40√3 + 15/2, -20 + 15√3/2)
Step 10: Find the magnitude of -A--> + 3B--> using the Pythagorean theorem.
Magnitude (|-A + 3B|) = √((-40√3 + 15/2)^2 + (-20 + 15√3/2)^2)
Note: The direction of each resultant vector can be found by using the arctan function to determine the angle it makes with the positive x-axis.
To find the magnitude and direction of A--> - B-->, we need to first determine the x- and y-components of each vector, and then subtract the corresponding components.
Given:
Magnitude of A--> (|A|) = 20m
Angle of A--> above the positive x-axis (θ) = 30 degrees
Magnitude of B--> (|B|) = 15m
Angle of B--> to the left of the y-axis (φ) = 60 degrees
1. Finding the x- and y-components of A-->:
We can use trigonometry to find the x and y component of A-->.
Ax = |A| * cos(θ) = 20m * cos(30 degrees) = 20m * √3/2 = 10√3 m
Ay = |A| * sin(θ) = 20m * sin(30 degrees) = 20m * 1/2 = 10m
2. Finding the x- and y-components of B-->:
Similarly, we can find the x and y component of B--> using trigonometry.
Bx = 0m (since it is oriented 60 degrees to the left of the y-axis)
By = |B| * sin(φ) = 15m * sin(60 degrees) = 15m * √3/2 = 15√3/2 m
3. Subtracting the components:
To find the x and y component of A--> - B-->, we subtract the corresponding components:
Ax - Bx = 10√3 m - 0m = 10√3 m
Ay - By = 10m - 15√3/2 m = 10m - 7.5√3 m
4. Finding the magnitude and direction of A--> - B-->:
Magnitude of A--> - B--> (|A - B|) = √((Ax - Bx)^2 + (Ay - By)^2)
|A - B| = √((10√3)^2 + (10 - 7.5√3)^2) ≈ √(300 + 225) ≈ √525 ≈ 22.91 m
To find the direction, we can use the inverse tangent function:
Direction of A--> - B--> (θ) = tan^(-1)((Ay - By) / (Ax - Bx))
θ = tan^(-1)((10 - 7.5√3) / (10√3)) ≈ 18.43 degrees
Therefore, the magnitude of A--> - B--> is approximately 22.91m, and its direction is approximately 18.43 degrees above the positive x-axis.
5. Finding the magnitude and direction of 2A--> + B-->:
To find the magnitude and direction of 2A--> + B-->, we need to double the components of A--> and add them to the components of B-->, then use the same magnitude and direction formulas as above.
Magnitude of 2A--> + B--> (|2A + B|) = √((2Ax + Bx)^2 + (2Ay + By)^2)
Direction of 2A--> + B--> (θ) = tan^(-1)((2Ay + By) / (2Ax + Bx))
6. Finding the magnitude of -A--> + 3B-->:
To find the magnitude of -A--> + 3B-->, we need to multiply the components of A--> by -1, multiply the components of B--> by 3, add the corresponding components, and then use the magnitude formula as above.
Magnitude of -A--> + 3B--> (|-A + 3B|) = √((-Ax + 3Bx)^2 + (-Ay + 3By)^2)
To find the magnitude and direction of A--> - B-->, we first need to find the components of both vectors A and B.
For vector A, the x-component can be found using cos(30°) = adjacent/hypotenuse = x/20m.
x = 20m * cos(30°) = 17.32m.
The y-component of vector A can be found using sin(30°) = opposite/hypotenuse = y/20m.
y = 20m * sin(30°) = 10m.
So, vector A--> can be represented as A--> = <17.32m, 10m>.
For vector B, we need to find the x and y-components given the orientation of 60 degrees to the left of the y-axis.
The angle between the y-axis and the vector B is 90° - 60° = 30°.
The x-component of vector B is then 15m * cos(30°) = 15m * 0.866 = 12.99m.
The y-component of vector B is -15m * sin(30°) = -15m * 0.5 = -7.5m (negative because it is oriented downward from the y-axis).
So, vector B--> can be represented as B--> = <12.99m, -7.5m>.
To find the magnitude and direction of A--> - B-->, we subtract the corresponding components:
A--> - B--> = <17.32m - 12.99m, 10m - (-7.5m)> = <4.33m, 17.5m>.
The magnitude of A--> - B--> is √(4.33^2 + 17.5^2) = 18.07m.
To find the direction, we use tan^-1(17.5m/4.33m) = 76.33° above the positive x-axis.
Therefore, the magnitude of A--> - B--> is 18.07m and the direction is 76.33° above the positive x-axis.
To find the magnitude and direction of 2A--> + B-->, we multiply vector A--> by 2 and then add vector B-->:
2A--> + B--> = 2<17.32m, 10m> + <12.99m, -7.5m> = <34.64m, 20m> + <12.99m, -7.5m> = <47.63m, 12.5m>.
The magnitude of 2A--> + B--> is √(47.63^2 + 12.5^2) = 49.27m.
To find the direction, we use tan^-1(12.5m/47.63m) = 15.1° above the negative x-axis.
Therefore, the magnitude of 2A--> + B--> is 49.27m and the direction is 15.1° above the negative x-axis.
To find the magnitude of -A--> + 3B-->, we multiply vector A--> by -1 and vector B--> by 3 and then subtract:
-A--> + 3B--> = -1<17.32m, 10m> + 3<12.99m, -7.5m> = <-17.32m, -10m> + <38.97m, -22.5m> = <21.65m, -32.5m>.
The magnitude of -A--> + 3B--> is √(21.65^2 + (-32.5)^2) = 39.16m.
Therefore, the magnitude of -A--> + 3B--> is 39.16m.