SOLUTION: the vector u has magnitude 10 and bearing 200 degrees. vector v has magnitude 8 and bearing 35 degrees. choose the correct magnitude of u-v and its bearings.

Question 35500: the vector u has magnitude 10 and bearing 200 degrees. vector v has magnitude 8 and bearing 35 degrees. choose the correct magnitude of u-v and its bearings.
Found 2 solutions by narayaba, venugopalramana:
Answer by narayaba(40) (Show Source): You can put this solution on YOUR website!
Vectors can be represneted in two possible ways. A vector can be represented using the cartesian cordinate system (x,y). They can also be represented as polar cordinates. In polar cordinate a vector is represented by its magnitude and its angle with respect to the positive direction of X axis.
If T is a vector with components (tx,ty) in cartesian cordinate system then in polar cordinate the vector can be written as follows
magnitude r = sqrt(tx^2 + ty^2) and angle theta = tan^-1(ty/tx) (inverse tan)
here tx = r*cos(theta) and ty = r*sin(theta) ------ EQUATION 1
now the vector T can be represented in polar cordinate as (r,theta)
For the vector u with magnitude r =8 and theta =35
the x and y components are ux = 8*cos(35) = 8* 0.8192 = 6.5536
and uy =8*sin(35) = 8* 0.5736 = 4.5586
similarly For the vector v with magnitude r =10 and theta =200
the x and y components are vx = 10*cos(200) = 10* -0.9397 = -9.9397
and vy =10*sin(200) = 10*-0.3420 = -3.420
we want vector u - v in cartesian coordinates (ux-vx, uy-vy) = (16.4933, 7.9786)
Using EQUATION 1
The vector u - v in polar cordinates is r = sqrt(16.4933^2 + 7.9786^2) = 18.3218
and theta = tan^-1(7.9786/16.4933) = 25.8153 degrees
The vector u - v in polar coordinates is (18.3218, 25.8153)

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
the vector u has magnitude 10 and bearing 200 degrees. vector v has magnitude 8 and bearing 35 degrees. choose the correct magnitude of u-v and its bearings.
ONE STANDARD METHOD IS TO DECOMPOSE EACH VECTOR INTO ITS COMPONENTS AND THEN ADD OR SUBTRACT OR WHATEVER IS NEEDED
U=10COS(200)*i+10SIN(200)*j=-9.403i-3.404j
V=8COS(35)*i+8SIN(35)*j=6.555i+4.587j
U-V=-15.9576i-7.99j
|U-V|=SQRT{(-15.9576)^2+(-7.99)^2}=17.84622
TAN(BEARING)=0.5008
BEARING=26.6 DEGREES
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