Question

What are the magnitude and direction of a vector that has an x component of -38.0 units and a y component of -60.0 units?

Answer 1

The magnitude of the vector is 71.0 units and the direction is 56.2 degrees south of the x-axis.

Step-by-step explanation:

To find the magnitude and direction of a vector with given x and y components, we can use the Pythagorean theorem and trigonometry. The magnitude of the vector can be found using the formula: magnitude = √(x^2 + y^2). Plugging in the given values, we have magnitude = √((-38.0)^2 + (-60.0)^2) = √(1444 + 3600) = √5044 = 71.0 units.

The direction of the vector can be found using the formula: direction = arctan(y / x). Plugging in the given values, we have direction = arctan((-60.0) / (-38.0)) = arctan(1.579) = 56.2 degrees south of the x-axis.

Answer 2

The magnitude of the vector is 71.04 units and its direction is 58.69 degrees north of the x-axis.

Step-by-step explanation:

The magnitude of a vector can be found using the Pythagorean theorem, which states that the magnitude (or length) of a vector is equal to the square root of the sum of the squares of its components. In this case, the x component is -38.0 units and the y component is -60.0 units.

So the magnitude of the vector is sqrt((-38.0)^2 + (-60.0)^2) = sqrt(1444 + 3600) = sqrt(5044) = 71.04 units.

To find the direction of the vector, you can use trigonometry. The angle can be found using the inverse tangent function (tan^-1) with the ratio of the y component to the x component. In this case, the ratio is -60.0 / -38.0 = 1.5789.

So the direction of the vector is atan(1.5789) = 58.69 degrees (or north of the x-axis).