If the expression for a given vector is Vector A-Vector B=(4-5)i^+(3+2)j^ and the magnitude of the difference of the vector is equal to 5.1. What is the direction of the difference Vector A - Vector B. (in the counterclockwise and +x direction)
To find the direction of the difference vector A - B, we need to determine the angle it makes with the positive x-axis in the counterclockwise direction.
First, let's rewrite the given expression for the difference vector A - B in its component form: (4 - 5)i + (3 + 2)j.
The magnitude of the difference vector is given as 5.1. We can use this information to find the magnitude of the vector in terms of its x and y components.
Magnitude of a vector P = √(Px^2 + Py^2)
Given that the magnitude of the difference vector A - B is 5.1, we can write:
√((4 - 5)^2 + (3 + 2)^2) = 5.1
Simplifying the equation, we have:
√((-1)^2 + 5^2) = 5.1
√(1 + 25) = 5.1
√26 = 5.1
Now, we know the magnitude of the difference vector is √26.
To find the direction of the difference vector, we can use the definition of the tangent function:
tan(θ) = Py / Px
Where θ is the angle and Py and Px are the y and x components of the vector, respectively.
Given the components as (4 - 5)i + (3 + 2)j, we can determine the direction:
tan(θ) = (3 + 2) / (4 - 5)
tan(θ) = 5 / (-1)
Now, we can find the angle θ by taking the inverse tangent (arctan) of both sides:
θ = arctan(5 / (-1))
Using a calculator, the angle in radians is approximately -1.3734 radians.
Since we want the direction in the counterclockwise direction, we need to add 2π (the equivalent of 360°) to the angle, giving us:
θ = -1.3734 + 2π
Finally, convert the angle to degrees, if needed, by multiplying by 180/π:
θ = (-1.3734 + 2π) * 180/π
Now, calculate the final value of θ using the desired units.