Question

Vector A is 5.5cm long and points along the x-axis. Vector B is 7.5cm long and points at +30 degrees above the negative x axis.

a. Determine the x and y components of vector A and vector B
b. Determine the sum of these two vectors in terms of components.
c. Determine the sum of these two vectors in terms of magnitude and direction

Answer 1

a)
`A_x=5.5cm\\A_y=0cm\\B_x=-6.495cm\\B_y=3.75cm\\`

b)
`(A+B)_x=-0.995cm\\(A+B)_y=3.75cm`

c)
`|A+B|=3.880cm \\ \theta_(A+B)=104.86^(\circ)`

Explanation:

If we call A and B the longitude of the vectors, and measure
`\theta` from the positive x axis in anti-clockwise direction (as usual), the x and y components are determined by the formulas:

`A_x=Acos(\theta_A)\\A_y=Asin(\theta_A)\\B_x=Bcos(\theta_B)\\B_y=Bsin(\theta_B)\\`

An angle of 30 degrees above the negative x axis is an angle of 180-30=150 degrees from the positive x axis in anti-clockwise direction. So we will have:

`A_x=(5.5cm)cos(0^(\circ))=5.5cm\\A_y=(5.5cm)sin(0^(\circ))=0cm\\B_x=(7.5cm)cos(150^(\circ))=-6.495cm\\B_y=(7.5cm)sin(150^(\circ))=3.75cm\\`

The sum of the 2 vectors (each of the components of their sum) in terms of their components will be:

`(A+B)_x=A_x+B_x=5.5cm-6.495cm=-0.995cm\\(A+B)_y=A_y+B_y=0cm+3.75cm=3.75cm`

Since
`(A+B)_x` and
`(A+B)_y` are the sides of a right triangle, the magnitude of the sum of A+B (which we will call |A+B|) will be given by the Pythagoras formula, and the angle can be calculated with the formula of the arc tangent since
`tan \theta_(A+B)=((A+B)_y)/((A+B)_x)`:

`|A+B|=√((A+B)_x^2+(A+B)_y^2)=√((-0.995cm)^2+(3.75cm)^2)=3.880cm`

For the angle one has to be careful since the arctan function cannot interpret some results. In our case, the vector A+B is on Quadrant II (negative x, positive y), so the best way will be to calculate the angle above negative x (using the absolute values of the components), and to 180 substract this value, then:

`\theta_(A+B)=180^(\circ)-arctan({(|(A+B)_y|)/(|(A+B)_x|))=180^(\circ)-arctan(3.76884422111)=104.86^(\circ)`