Question

The magnitude and direction of two vectors are shown in the diagram. What is the magnitude of their sum? ​

Answer 1

Answer: 2√5

Step-by-step explanation:

Use the Sine Law of triangles: In triangle with sides of length 1,2,3 that are opposite to angles 1,2,3 we have

1/sin1=2/sin2=3/sin3.

Answer 2

Separate the vectors into their x- and y-components. Let u be the vector on the right and v the vector on the left, so that

u = 4 cos(45°) x + 4 sin(45°) y

v = 2 cos(135°) x + 2 sin(135°) y

where x and y denote the unit vectors in the x and y directions.

Then the sum is

u + v = (4 cos(45°) + 2 cos(135°)) x + (4 sin(45°) + 2 sin(135°)) y

and its magnitude is

||u + v|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)

… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))

… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))

… = √(16 + 16 cos(135° - 45°) + 4)

… = √(20 + 16 cos(90°))

… = √20 = 2√5