a) The horizontal component of the resultant vector is equal to the sum of the horizontal components of vector à and vector B. Since vector B is directed along the positive x-axis, its horizontal component is equal to its magnitude, which is 4 units. Vector à makes an angle of 30° with the positive x-axis, so its horizontal component can be calculated using cosine:
Horizontal component of à = 4 * cos(30°) = 4 * √3/2 = 2√3
Therefore, the horizontal component of the resultant vector is 2√3 units.
b) Since vector B is directed along the positive x-axis, it does not have a vertical component. The vertical component of vector à can be calculated using sine:
Vertical component of à = 4 * sin(30°) = 4 * 1/2 = 2
Therefore, the vertical component of the resultant vector is 2 units.
c) To calculate the magnitude of the resultant vector, we can use the Pythagorean theorem:
Magnitude of the resultant vector = √(horizontal component)^2 + (vertical component)^2
= √(2√3)^2 + 2^2
= √(12) + 4
= √16
= 4 units
Therefore, the magnitude of the resultant vector is 4 units.
d) The direction of the resultant vector can be calculated using the arctan function:
Direction of the resultant vector = arctan(vertical component / horizontal component)
= arctan(2 / 2√3)
= arctan(1 / √3)
= 30°
Therefore, the direction of the resultant vector is 30° above the positive x-axis.
5. Vector à has magnitude of 4 units and makes an angle of 30° with the positive x-axis. B
Vector also has the same magnitude of 4 units and directed along the positive x-axis.
Calculate:
a) the horizontal component of the resultant vector,
b) vertical component of the resultant vector,
c) the magnitude of the resultant vector,
d) the direction of the resultant vector.
5 answers
wrong again!
If B≠kA , then A+B cannot have the same direction as A!
If B≠kA , then A+B cannot have the same direction as A!
You are correct, I apologize for the mistake. If B is not a scalar multiple of A, then A+B cannot have the same direction as A.
In this case, since vector B is directed along the positive x-axis, it has no vertical component. The horizontal component of vector à can be calculated using cosine:
Horizontal component of à = 4 * cos(30°) = 4 * √3/2 = 2√3
Therefore, the horizontal component of the resultant vector is 2√3 units.
The vertical component of vector à can be calculated using sine:
Vertical component of à = 4 * sin(30°) = 4 * 1/2 = 2
Therefore, the vertical component of the resultant vector is 2 units.
To calculate the magnitude of the resultant vector, we can use the Pythagorean theorem:
Magnitude of the resultant vector = √(horizontal component)^2 + (vertical component)^2
= √(2√3)^2 + 2^2
= √12 + 4
= √16
= 4 units
Therefore, the magnitude of the resultant vector is 4 units.
To find the direction of the resultant vector, we can use trigonometric functions:
Direction of the resultant vector = arctan(vertical component / horizontal component)
= arctan(2 / 2√3)
= arctan(1 / √3)
= 30°
Therefore, the direction of the resultant vector is 30° above the positive x-axis.
In this case, since vector B is directed along the positive x-axis, it has no vertical component. The horizontal component of vector à can be calculated using cosine:
Horizontal component of à = 4 * cos(30°) = 4 * √3/2 = 2√3
Therefore, the horizontal component of the resultant vector is 2√3 units.
The vertical component of vector à can be calculated using sine:
Vertical component of à = 4 * sin(30°) = 4 * 1/2 = 2
Therefore, the vertical component of the resultant vector is 2 units.
To calculate the magnitude of the resultant vector, we can use the Pythagorean theorem:
Magnitude of the resultant vector = √(horizontal component)^2 + (vertical component)^2
= √(2√3)^2 + 2^2
= √12 + 4
= √16
= 4 units
Therefore, the magnitude of the resultant vector is 4 units.
To find the direction of the resultant vector, we can use trigonometric functions:
Direction of the resultant vector = arctan(vertical component / horizontal component)
= arctan(2 / 2√3)
= arctan(1 / √3)
= 30°
Therefore, the direction of the resultant vector is 30° above the positive x-axis.
you came up with the same wrong answer!
A = <2√3,2>
B = <4,0>
so A+B = <4+2√3,2> with magnitude = 7.727
so tanθ = 2/7.464 and θ = 15°
A = <2√3,2>
B = <4,0>
so A+B = <4+2√3,2> with magnitude = 7.727
so tanθ = 2/7.464 and θ = 15°
Apologies for the confusion. You are correct.
A = <2√3, 2>
B = <4, 0>
A + B = <2√3 + 4, 2>
The horizontal component of the resultant vector A + B is 2√3 + 4 units.
The vertical component of the resultant vector A + B is 2 units.
To calculate the magnitude of the resultant vector, we can use the Pythagorean theorem:
Magnitude of the resultant vector = √((2√3 + 4)^2 + 2^2)
= √(12 + 16√3 + 12 + 4)
= √(28 + 16√3)
= 7.727 units
The direction of the resultant vector can be calculated using the arctan function:
Direction of the resultant vector = arctan(vertical component / horizontal component)
= arctan(2 / (2√3 + 4))
= arctan(2 / 7.464)
≈ 15°
Therefore, the magnitude of the resultant vector is approximately 7.727 units and its direction is approximately 15° above the positive x-axis.
A = <2√3, 2>
B = <4, 0>
A + B = <2√3 + 4, 2>
The horizontal component of the resultant vector A + B is 2√3 + 4 units.
The vertical component of the resultant vector A + B is 2 units.
To calculate the magnitude of the resultant vector, we can use the Pythagorean theorem:
Magnitude of the resultant vector = √((2√3 + 4)^2 + 2^2)
= √(12 + 16√3 + 12 + 4)
= √(28 + 16√3)
= 7.727 units
The direction of the resultant vector can be calculated using the arctan function:
Direction of the resultant vector = arctan(vertical component / horizontal component)
= arctan(2 / (2√3 + 4))
= arctan(2 / 7.464)
≈ 15°
Therefore, the magnitude of the resultant vector is approximately 7.727 units and its direction is approximately 15° above the positive x-axis.