a vector has a magnitude of 16 and direction of 120 degrees. write the vector in terms of unit vecotrs i and j.
16 cos 120 i + 16 sin 120 j
cos 120 = -.5
sin 120 = +.866 or (1/2)sqrt 3
so
-8 i + 8 sqrt 3 j
SKETCH it on a GRAPH !!!!
To write the vector in terms of unit vectors i and j, we can use trigonometry and the concept of components.
First, let's visualize the situation. We have a vector with a magnitude of 16 and a direction of 120 degrees. This means the vector forms an angle of 120 degrees with the positive x-axis in the counterclockwise direction.
To express the vector in terms of unit vectors i and j, we need to determine the components of the vector in the x and y directions.
The magnitude of the vector can be represented using the Pythagorean theorem:
Magnitude^2 = x-component^2 + y-component^2
Given the magnitude is 16, we have:
16^2 = x-component^2 + y-component^2
256 = x-component^2 + y-component^2
Now, let's determine the components using trigonometry.
Since the given angle is 120 degrees, we can draw a right triangle where the hypotenuse represents the magnitude of the vector and the opposite and adjacent sides represent the y and x components, respectively.
Using trigonometry, we know that:
cosine(angle) = adjacent / hypotenuse
sine(angle) = opposite / hypotenuse
We can rewrite these equations to solve for the components:
adjacent = hypotenuse * cosine(angle)
opposite = hypotenuse * sine(angle)
Plugging in the given values:
x-component = 16 * cosine(120)
y-component = 16 * sine(120)
To simplify further, we can use the fact that cosine(120) = -0.5 and sine(120) = √3 / 2:
x-component = 16 * (-0.5) = -8
y-component = 16 * (√3 / 2) = 8√3
Finally, we can write the vector in terms of unit vectors i and j:
Vector = -8i + 8√3j