Vector C has a magnitude of 10 units and is dir... - SS2 Physics Mechanics - Motion in a Plane Question

To find the vector difference C - D, we need to subtract the x-components and the y-components of the vectors separately.

 

For vector C:

Cx = 10 x cos(45°) = 10 x 0.707 = 7.071 units

Cy = 10 x sin(45°) = 10 x 0.707 = 7.071 units

 

For vector D:

Dx = 6 x cos(330°) = 6 x 0.866 = 5.196 units

Dy = 6 x sin(330°) = 6 x (-0.5) = -3 units

 

Now, subtract the x-components and y-components separately:

Difference of x-components = Cx - Dx = 7.071 - 5.196 = 1.875 units

Difference of y-components = Cy - Dy = 7.071 - (-3) = 10.071 units

 

To find the magnitude of the vector difference, use the Pythagorean theorem:

Magnitude = sqrt((Difference of x-components)2 + (Difference of y-components)2)

Magnitude = ((1.875)2 + (10.071)2) ≈ (3.515625 + 101.429641) ≈ (104.945266) ≈ 10.244 units

 

To find the direction of the vector difference, use the inverse tangent function:

Direction = tan(-1)((Difference of y-components) / (Difference of x-components))

Direction = tan(-1)(10.071 / 1.875) ≈ 80.9 degrees

 

Therefore, the magnitude of the vector difference C - D is approximately 10.244 units, and its direction is approximately 80.9 degrees above the positive x-axis.