Trigonometry Vector problems

Question

For the vector 𝑎 = 12 units at 𝜃 = 160° and 𝑏 = 8 units at 𝜃 = 310° , find 𝑎 + 𝑏

1.As the sum of two components

2. As a magnitude and direction

Answer 1

1.

a =〈12cos(160°),12sin(160°)〉≈〈-11.27,4.10〉

b = 〈8cos(310°),8sin(310°)〉≈〈5.14,-6.13〉

a+b ≈〈-6.13, -2.02〉

2.

|a+b| = √[(-6.13)²+(-2.02)²] ≈ 6.46

θ ≈ tan^-1(-2.02/-6.13) ≈ 18.26°+180° = 198.26° (third quadrant)

Answer 2

Drawing these vectors we get:

The components would be:

a_x = 12cos(160°) = -11.2763

a_y = 12sin(160°) = 4.1042

b_x = 8cos(310°) = 5.1423

b_y = 8sin(310°) = -6.1284

The resultant vector (R) will have components:

R_x = a_x + b_x = -11.2763 + 5.1423 = -6.1340

R_y = a_y + b_y = 4.1042 - 6.1284 = -2.0241

The magnitude of the resultant vector R is √[(-6.1340)² + (-2.0241)²] = 6.459

The angle (because both R_x and R_y are negative will be in quadrant 3. We use tan(θ) = y/x = -2.0241/-6.1340 so θ = tan^-1(-2.0241/-6.1340) = 18.26° however, to get this angle in quadrant 3, we need to add 180° to it to get θ = 198.26°

Answer 3

Emily:

Break into components. Remember a_x = a * cos(θ) and a_{y =} a*sin(θ). Similarly, b_x = b*cos(θ) and b_y = b*sin(θ)

We know a_x = 12*cos(160) = -11.27 a_y = 12*sin(160) = 4.1

b_x = 8*cos(310) = 5.1423 b_y = 8*sin(310) = -6.128

We can then add the components together

Let c = a + b

c_x = a_x + b_x = -11.27 + 5.1423 = -6.1277

c_{y =} a_{y +} b_y = 4.1 -6.128 = -2.028

So c = -6.1277*i + -2.028j

tan(θ) = -2.028/-6.1277 remember to get the right quadrant (where both x and y values are negative 180<θ<270)

Let me know if you have any other questions!