Vector A is directed along the positive y-axis and has magnitude sqrt 3.0 units. Vector B is directed along the negative x-axis and has a magnitude 1.0 units. (a) what are the magnitude and direction of A-B? (b) what are the x and y components of B-A?
To find the magnitude and direction of A-B, we need to subtract B from A.
(a) Magnitude of A-B:
To subtract vectors, we need to reverse the direction of the vector we are subtracting and then add them together. In this case, we want to find A-B.
Vector A is directed along the positive y-axis with a magnitude of √3.0 units. Since we want to subtract vector B, we need to reverse its direction, which means B will be directed along the positive x-axis with a magnitude of 1.0 units.
Now, we can subtract the two vectors:
A-B = A + (-B)
From the given information, the magnitude of A is √3.0 units and the magnitude of B is 1.0 unit.
Using the Pythagorean theorem, the magnitude of A-B can be found as follows:
Magnitude of A-B = √[(√3.0)^2 + 1.0^2]
= √[3.0 + 1.0]
= √4.0
= 2.0 units
Therefore, the magnitude of A-B is 2.0 units.
Direction of A-B:
To find the direction of A-B, we can use trigonometry. Since vector A is directed along the positive y-axis and vector B is directed along the positive x-axis, vector A-B will be in the second quadrant.
Using the arctan function, we can find the angle:
Angle = arctan(y-component / x-component)
= arctan(√3.0 / -1.0)
= arctan(-√3.0)
≈ -60 degrees
Therefore, the direction of A-B is approximately -60 degrees (measured counterclockwise from the positive x-axis).
(b) To find the x and y components of B-A, we need to reverse the order of subtraction (A-B) and then find the components.
Reversing the order, B-A = -1 * (A-B)
The x component of B-A can be found as follows:
x-component of B-A = -1 * √3.0
= -√3.0
The y component of B-A can be found as follows:
y-component of B-A = -1 * 1.0
= -1.0
Therefore, the x and y components of B-A are -√3.0 and -1.0, respectively.