Subtracting vectors

Question

If vector 1 has a magnitude of 3 units and a direction of 30 degrees below the negative x-axis, and vector 2 has a magnitude of 4 units and a direction of 80 degrees below the negative x-axis, what is vector 1 minus vector 2?

Please complete this question geometrically using images

Answer 1

Here is a sketch:

See how I've taken the two vectors and drawn separate triangles for them? Use trig ratios to solve for the sides:

For Vector 1:

sin(θ) = opp/hyp

sin(30°) = V1_y/3

V1_y = 3sin(30°)

V1_y = 1.5 but, since it is pointed downward, we make V1_y = -1.5

And cos(θ) = adj/hyp

cos(30°) = V1_x/3

V1_x = 3cos(30°)

V1_x = 2.598 but, since it is pointed left, we make V1_x = -2.598

For Vector 2:

sin(θ) = opp/hyp

sin(80°) = V2_y/4

V2_y = 4sin(80°)

V2_y = 3.939 but, since it is pointed downward, we make V2_y = -3.939

And cos(θ) = adj/hyp

cos(80°) = V2_x/4

V2_x = 4cos(80°)

V2_x = 0.695 but, since it is pointed left, we make V2_x = -0.695

Now subtract: V1 - V2:

First do the x-direction:

V1_x - V2_x

-2.598 - -0.695 = -2.598 + 0.695 = -1.903

Then do the y-direction:

V1_y - V2_y

-1.5 - -3.939 = -1.5 + 3.939 = 2.439

So the result has an x-direction component of -1.903 and a y-direction component of 2.439. To determine the hypotenuse (which is the vector (V1 - V2) we first find its magnitude or length using the Pythagorean Theorem:

magnitude = √[(-1.903)² + (2.439)²] = √(3.623 + 5.950) = √9.573 = 3.094

To find the direction of the vector first think about where it is considering the x-direction component is -1.903 and the y-direction component is 2.439. Can you see that it is in quadrant 2? The angle above the negative x-axis would be found using tan^-1(2.439/1.903) = 52.033° Notice that I used a positive number for 1.903. This was intentional because I am using the reference angle and know the result will be ABOVE the negative x-axis in Q2.

Answer 2