A vector has an x component of -30.0 units and a y component of 49.5 units. Find the magnitude and direction of the vector.
To find the magnitude of the vector, we can use the Pythagorean theorem. The magnitude (or length) of a vector is given by:
Magnitude = sqrt(x^2 + y^2)
Substituting the given values, we have:
Magnitude = sqrt((-30.0)^2 + (49.5)^2)
= sqrt(900 + 2450.25)
= sqrt(3350.25)
≈ 57.9 units (rounded to one decimal place)
So, the magnitude of the vector is approximately 57.9 units.
To find the direction of the vector, we can use trigonometry. The angle θ between the positive x-axis and the vector is given by:
θ = arctan(y / x)
Substituting the given values, we have:
θ = arctan(49.5 / -30.0)
≈ -59.5 degrees (rounded to one decimal place)
Note that the negative sign indicates that the direction is in the 3rd quadrant (below the x-axis).
Therefore, the magnitude of the vector is approximately 57.9 units and its direction is approximately -59.5 degrees (counterclockwise from the positive x-axis).
To find the magnitude and direction of the vector, we can use the Pythagorean theorem and trigonometry.
Step 1: Calculate the magnitude of the vector using the Pythagorean theorem:
Magnitude = √(x^2 + y^2)
Magnitude = √((-30.0)^2 + (49.5)^2)
Magnitude = √(900 + 2450.25)
Magnitude = √(3350.25)
Magnitude ≈ 57.91 units
Step 2: Calculate the direction of the vector using trigonometry:
Direction = arctan(y / x)
Direction = arctan(49.5 / -30.0)
Direction ≈ -57.24 degrees
Therefore, the magnitude of the vector is approximately 57.91 units and the direction is approximately -57.24 degrees.
With the number of questions on vectors you have asked lately, you would certainly be able to work out part of the question, or else you will have a problem at exam time.
Here is how you can do it. If you have difficulties, post what you get.
Vector(-30, 49.5)
Magnitude=√(x²+y²)
=√((-30)²+49.5²)
=?
The direction is a little more tricky, as you will have to determine which quadrant the vector belongs.
Since x<0 and y>0, we determine that it is in the second quadrant.
Find θ=tan-1(49.5/30) and subtract from 180° (2nd quadrant) to get the direction of the vector.