Question

Vector A has magnitude 8.00 mm and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axis). The sum A +B is in the −y-direction and has magnitude 12.0 mm. Find the vector B.

Answer 1

Answer:

-75.35°

Explanation:

Let C be the sum of the two vectors A and B. Hence, we can write the following  

$A_{x} +B_{x} =C_{x} ......(1)\\A_{y} +B_{y} =C_{y} ......(2)$

but since the vector C is in the -y direction, $C_{x}$  = 0 and $C_{y}$ = —12 m.  

Thus  

$B_{x} =-A_{x} =-[-Acos(180-127)]=(8)*cos(53)\\B_{x} =4.81m$

similarly, we can determine $B_{y}$ by rearranging equation (1)  

 $B_{y} =C_{y} -A_{y} =-12m-[(8)*sin(53)\\B_{y} =-18.4m$

so the magnitude of B is

$B=\sqrt{B_{x}^2+B_{y}^2 } \\B=19m$

Finally, the direction of B can be calculated as follows  

Ф=$tan^{-1} (\frac{B_{y} }{B_{x} } )\\=-75.35$

hence the vector B makes an angle of 75.35 clockwise with + x axis