Question

Vector A has a magnitude of 8.00 units and makes an angle of 45.0° with the positive x-axis. Vector B also has a magnitude of 8.00 units and is directed along the negative x-axis. Find (a) the vector sum A + B and (b) the vector sum of A-B?

Answer 1

Answer:

1) $\bar A + \bar B = -2.34 i + 5.66 j$

2) $\bar A - \bar B = 13.66 i + 5.66 j$

Step-by-step explanation:

Given:

IAI = 8.00 units  directed at angle of 45.0 degrees with +x - axis

IBI = 8.00 units  directed along -x - axis.

Components of vector along x and y directions:

$A_x = (8.00) (cos(45.0^0)) = 5.66\\A_y = (8.00)(sin(45.0^0)) = 5.66$

$B_x = (8.00)(cos(180^0)) = -8.00\\B_y = (8.00)(sin(180^0)) = 0$

$\bar A = (A_x) i + (A_y) j = 5.66 i + 5.66 j\\\bar B = (B_x) i + (B_y) j = -8.00 i + 0 j$

1) To find Vector sum $( \bar A + \bar B)$

$(\bar A + \bar B) = (5.66-8) i + ( 5.66+0) j = -2.34 i + 5.66 j$

2) To find Vector sum $(\bar A - \bar B)$

$(\bar A - \bar B) = (5.66-(-8)) i + ( 5.66-0) j = -13.66 i + 5.66 j$