Please help me with these questions??

Question

Please help me (the two Questions are seperated, the second question has "a" and "b")

1.

Two vectors A and B have magnitude A= 3.00 and B= 3.02

Their vector product is A x B = -5.06 k + 1.91 i

What is the angle between A and B
-----------------------------------------------------------------------------------
2.

(see the picture for question number 2)

http://img100.imageshack.us/my.php?image=joo9.jpg


a.

If R is the sum of three vectors (A,B,C) and it equals 16.6 m, find it's direction (take an angle positive counterclockwise from the + x-axis)

b. Find the magnitude of the vector S=C-A-B, and it's direction

thanks a lot in advance

Answer 1

Use the definition

Part a
By definition |A x B|=|A| |B| sin(a) , where a in the angle between A and B

Since |A|= 3.00 and |B|= 3.02
And
A x B = -5.06 k + 1.91 i then
|AxB|=-sqrt( (-5.06)^2 + (1.91)^2)=5.408
Then sin(a)= |AxB|/(A| |B|)
sin(a)= 5.408/(3.00)(3.02))= .597
a=arcSin(.597)=36.65 degrees

Part b
A = 12m angle (90-37)=53 degrees
B = 15m angle -40 degrees (or 320)
C = 5m angle (180 + 60)= 240 degrees
Other angle conventions are possible this way it is easier to use the same definition.
A=Axi+Ayj where
Ax=A cos(a)
Ay=A sin(a) same is true for the rest of the vectors

And R=(Ax+Bx+Cx)i + (Ay+By+Cy)j
The magnitude of |R|==sqrt((Ax+Bx+Cx)^2 + (Ay+By+Cy)^2)
Angle or r= arctan((Ay+By+Cy)/( Ax+Bx+Cx))

Ax= 12cos(53)=7.22m
Bx=15cos(-40)=11.49m
Cx = 5cos(240)=-2.5m
Rx=16.21m

Ay=12sin(53)=9.5m
By=15sin(-40)=-9.64m
Cy=5sin(240)=-4.33m
Ry=-4.47m

R=(16.21i – 4.47j)m
|R|= 16.82m
r=atan(-4.47/16.21)=-15.4 degrees (or 344.6)