A vector has an x-component of −21.0units and a y-component of 36.5 units. Find the magnitude and direction of the vector.
X = hor. = -21 Units.
Y = ver. = 36.5 Units.
tanA = y/X = 36.5 / -21 = -1.738,
A = -60 deg. = -60 + 180 = 120 deg.,CCW.
R = X / cosA = -21 / cos(120) = 42. @ -60 deg. = -60 + 180 = 120 deg. =CCW.
R = (42,120deg.).
To find the magnitude of the vector, we can use the Pythagorean theorem.
The magnitude is given by the formula:
Magnitude = √(x^2 + y^2)
Given values:
x-component = -21.0 units
y-component = 36.5 units
Substituting the values into the formula:
Magnitude = √((-21)^2 + 36.5^2)
Magnitude = √(441 + 1332.25)
Magnitude = √(1773.25)
Magnitude ≈ 42.10 units (rounded to two decimal places)
To find the direction, we can use the inverse tangent function.
Direction = arctan(y-component / x-component)
Substituting the values into the formula:
Direction = arctan(36.5 / -21.0)
Calculating the value using a calculator:
Direction ≈ -60.61 degrees (rounded to two decimal places)
Therefore, the magnitude of the vector is approximately 42.10 units, and the direction is approximately -60.61 degrees.
To find the magnitude of the vector, you can use the Pythagorean theorem. The magnitude of a vector is the square root of the sum of the squares of its components.
Magnitude = √(x-component^2 + y-component^2)
In this case, the x-component is -21.0 units and the y-component is 36.5 units, so
Magnitude = √((-21.0)^2 + (36.5)^2)
= √(441 + 1332.25)
= √1773.25
≈ 42.08
The magnitude of the vector is approximately 42.08 units.
To find the direction of the vector, you can use trigonometry. The direction can be represented by the angle that the vector makes with the positive x-axis.
Direction = arctan(y-component / x-component)
In this case, the y-component is 36.5 units and the x-component is -21.0 units, so
Direction = arctan(36.5 / -21.0)
≈ -59.53°
The direction of the vector is approximately -59.53° (measured counterclockwise from the positive x-axis).