Vector A has a magnitude of 75.0 cm and points at 30° above the positive x-axis. Vector B has a magnitude of 25.0 cm and points along the negative x-axis. Vector C⃗ has a magnitude of 40.0 cm and points at 45° below the negative x-axis.

(a) Determine the x and y components of Vector A

(b) Determine the x and y components of Vector B .

(c) Determine the x and y components of Vector C .

(d) Determine x and y components of the sum of these three vectors.

(e) Determine the magnitude of the sum of these three vectors.

(f) Determine the direction of the sum of these three vectors.

(b) "points along the negative axis" in other words means that B makes an angle of 180° with the positive x-axis in the counterclockwise direction. In particular this tells you that B is parallel to the x-axis, so its y component would be zero.

B = (25.0 cm) (cos(180°) i + sin(180°) j)

B = (25.0 cm) (-1 i + 0 j)

B = (-25.0 i) cm

(c) If the negative x-axis corresponds to 180°, then 45° below this would make an angle of 180° + 45° = 225° with the positive x-axis.

C = (40.0 cm) (cos(225°) i + sin(225°) j)

C = (40.0 cm) (-1/√2 i - 1/√2 j)

C ≈ (-28.28 i - 28.28 j) cm

(d) The sum A + B + C has components that are the sums of the components of each of A, B, and C.

A + B + C ≈ (11.67 i + 9.22 j) cm

(e) The magnitude of the vector sum is

||A + B + C|| = √((11.67 cm)² + (9.22 cm)²) ≈ 14.87 cm

(f) Since both components of the vector sum are positive, you know that it's a vector that terminates in the first quadrant of the x,y-plane, so it makes an angle with the positive x-axis of

arctan(9.22/11.67) ≈ 38.3°