Find the magnitude and direction angle for each vector. Give the measure of the direction angle as an angle in [0,360].
<5sqrt2,-5sqrt2>
|v| = √(5√2)^2 + (5√2)^2) = √(50+50) = 10
tanθ = y/x = -5√2 / 5√2 = -1
θ = -45°
Plot the point and you can easily see that this is so.
To find the magnitude and direction angle for a vector, we can use the following formulas:
Magnitude:
The magnitude of a vector (a, b) is given by the formula:
|v| = sqrt(a^2 + b^2)
Direction Angle:
The direction angle of a vector (a, b) is given by the formula:
θ = atan2(b, a) [in radians]
where atan2(b, a) is the arctangent function that returns the angle whose tangent is the quotient of two specified numbers b and a.
Now, let's calculate the magnitude and direction angle for the given vector <5√2, -5√2>.
Magnitude:
|v| = sqrt((5√2)^2 + (-5√2)^2)
= sqrt(25*2 + 25*2)
= sqrt(50 + 50)
= sqrt(100)
= 10
Therefore, the magnitude of the vector <5√2, -5√2> is 10.
Direction Angle:
θ = atan2(-5√2, 5√2) [in radians]
Using a calculator or programming language to compute atan2 function, we get:
θ ≈ -45° (in degrees)
Since the direction angle, θ, is negative, we can add 360° to get a positive angle within the range of [0, 360]:
θ + 360° ≈ -45° + 360° ≈ 315°
Therefore, the direction angle for the vector <5√2, -5√2> is approximately 315°, measured counterclockwise from the positive x-axis.