Physics, Vectors

Question

Given the following vectors:  = 7.2î + 4.0ĵ,  = 5.0î + 4.3ĵ, and  = 6.4î + 5.9ĵ(a) Find the vector , in unit vector notation, such that  + 1 - 2 + 3 = 0. 

 =  

 

(b) Express your answer in part (a) in terms of magnitude and angle counterclockwise from the + x direction. (Indicate the direction with the sign of your answer.) 

 ? at   °

Answer 1

Vector addition is much like polynomial addition in that only like components can be added/subtracted together.

We see we have vectors:

1. A = 7.2*i + 4.0*j
2. B = 5.0*i + 4.3*j
3. C = 6.4*i + 5.9*j

Calculation of D must result in the following truthfulness of the equation...

D + 1*A - 2*B + 3*C = 0

1*A = A = 7.2*i + 4.0*j

-2*B = -10*i - 8.6*j

3*C = 19.2*i + 17.7*j

So D = -A + 2*B - 3*C

D = (-7.2 + 10 - 19.2)*i + (-4.0 + 8.6 - 17.7)*j = -16.4*i - 13.1*j

Calculation of the magnitude is to take the Pythagorian product (sqrt( sum of squares )). So let's do this like a polynomial...

sqrt(D*D) = ||D|| = magnitude

D*D = (-16.4*i - 13.1*j)(-16.4*i - 13.1*j) = 269.96 (i*i) + 214.84 (i*j) + 214.84 (j*i) + 171.61 (j*j)

Now, conveniently, i*j = j*i = 0. These represent "dot-products" and since they are 90 degrees from one another they multiply to 0. However, i*i = j*j = 1 (this is due to the fact the two "axes" align perfectly)

D*D = 269.96 + 171.61 = 441.57. (we had expected D*D to be positive and it is)

||D|| = sqrt(D*D) = 21.01 units. (This is the magnitude)

The rotation of vector D from the x-axis can be found via the tangent of the vector D (j-component/i-component)

tan(θ) = -13.1/-16.4 --> tan(θ) = .799

This yields an angle (θ) of 38.62 degrees, but that implies it is in the first quadrant and both the x (i) and y (j) values are negative, so this must be in the third quadrant and we need to add 180 degrees to our answer to place the vector into the proper quadrant.

Therefore θ = 218.62 degrees from the x-axis (counter-clockwise)