Resolving Vectors into Components – the “Vector Collector” Before adding the vectors, you should sketch the vectors. R = J + K J = 160.km [E25oN] K = 95.0km [SE] X Y Jx = Jy = Kx = Ky = Rx = Ry = J 25o 45o K

Resolving Vectors into Components – the “Vector Collector” Sketch the vector’s x & y components R = J + K J = 160.km [E25oN] K = 95.0km [SE] opp adj X Y Jx = Jy = Kx = Ky = Rx = Ry = J Jy 25o Jx K Ky 45o Kx

Resolving Vectors into Components – the “Vector Collector” Solve for the vector’s x & y components R = J + K J = 160.km [E25oN] K = 95.0km [SE] opp adj NOTE: sin 45 = cos 45 (Gotta those 45-45-90 triangles!) X Y Jx = (cos Ѳ) (hyp) = (cos 25o) (160 km) = 145 km [E] Jy = (sin Ѳ) (hyp) = (sin 25o) (160 km) = 67.6 km [N] Kx = (sin Ѳ) (hyp) = (sin 45o) (95.0 km) = 67.2km [E] Ky = (cos Ѳ) (hyp) = (cos 45o) (95.0 km) = 67.2km [S] Rx = Ry = J Jy 25o Jx K Ky 45o Kx

Resolving Vectors into Components – the “Vector Collector” R = J + K J = 160.km [E25oN] K = 95.0km [SE] opp adj adj add all the X’s to find the total X; then do the same for all the Y’s NOTE: sin 45 = cos 45 (Gotta those 45-45-90 triangles!) X Y Jx = (cos Ѳ) (hyp) = (cos 25o) (160 km) = 145 km [E] Jy = (sin Ѳ) (hyp) = (sin 25o) (160 km) = 67.6 km [N] Kx = (sin Ѳ) (hyp) = (sin 45o) (95.0 km) = 67.2km [E] Ky = (cos Ѳ) (hyp) = (cos 45o) (95.0 km) = 67.2km [S] Rx = 145 km [E] + 67.2 km [E] 212.2 km [E] 212 km [E] Ry = 67.6 km [N] + -67.2 km [S] [N] 0.4 km [N] J Jy 25o Jx K Ky 45o Kx

To re-assemble your vector, use pythagorean theorem R2 = Rx2 + Ry2 = (212)2 + (0.4)2 red number = 44944 + 0.16 denotes the R = √44944.16 last significant = 212 km [E Ѳ N] figure tanѲ = Ry so Ѳ = tan-1 Ry Rx Rx = tan-1 0.4 212 = 0.1o •• R = 212 km [E 0.1oN] Rx = 212 km [E] Ry = 0.4 km [N] R Rx Ѳ Ry •

Now try this one, R = P + Q P = 11.0 m [N27oE] Q = 15.0 m [E25oN] First step: Sketch your vectors (no peaking til everybody’s given it a try!!) X Y Px = Py = Qx = Qy = Rx = Ry =

R = P + Q P = 11.0 m [N27oE] Q = 15.0 m [E25oN] X Y Px = Py = Qx = Qy = Rx = Ry = Px Py 27o P Q Qy Now use the Vector Collector to calculate the components of each vector. 25o Qx

R = P + Q P = 11.0 m [N27oE] Q = 15.0 m [E25oN] X Y Px =(sin Ѳ) (hyp) = (sin 27o) (11.0 m) = 4.99 m [E] Py =(cos Ѳ) (hyp) = (cos 27o) (11.0 m) = 9.80 m [N] Qx =(cos Ѳ) (hyp) = (cos 25o) (15.0 m) = 13.6 m [E] Qy =(sin Ѳ) (hyp) = (sin 25o) (15.0 m) = 6.34 m [N] Rx = 4.99 m [E] + 13.6 m [E] 18.59 m [E] = 18.6 m [E] Ry = 9.80 m [N] + 6.34 m [N] 16.14 m [N] Px Py 27o P Q Qy 25o Now reassemble your resultant vector using the pythagorean theorem. Qx

R = P + Q R2 = Rx2 + Ry2 = (18.6)2 + (16.14)2 = 345.96 + 260.4996 R = √606.4596 = 24.6 m [E Ѳ N] Ѳ = tan-1 Ry Rx = tan-1 16.14 18.6 = 40.9o or 41 o R = 24.6 m [E41oN] X Y Px =(sin Ѳ) (hyp) = (sin 27o) (11.0 m) = 4.99 m [E] Py =(cos Ѳ) (hyp) = (cos 27o) (11.0 m) = 9.80 m [N] Qx =(cos Ѳ) (hyp) = (cos 25o) (15.0 m) = 13.6 m [E] Qy =(sin Ѳ) (hyp) = (sin 25o) (15.0 m) = 6.34 m [N] Rx = 4.99 m [E] + 13.6 m [E] 18.59 m [E] = 18.6 m [E] Ry = 9.80 m [N] + 6.34 m [N] 16.14 m [N] R Ry Ѳ Rx

How would it change if there was subtraction, R = P – Q How would it change if there was subtraction, R = P – Q ? Think “add the negative” R = P +(–Q) P = 11.0 m [N27oE] Q = 15.0 m [E25oN] -Q = 15.0 m [W25oS] “-Q” is the new name For the vector going in the exact opposite direction of “Q” X Y Px = Py = -Qx = -Qy = Rx = Ry =

R = P + (-Q) P = 11.0 m [N27oE] -Q = 15.0 m [W25oS] X Y Px =(sin Ѳ) (hyp) = (sin 27o) (11.0 m) = 4.99 m [E] Py =(cos Ѳ) (hyp) = (cos 27o) (11.0 m) = 9.80 m [N] -Qx =(cos Ѳ) (hyp) = (cos 25o) (15.0 m) = 13.6 m [W] -Qy =(sin Ѳ) (hyp) = (sin 25o) (15.0 m) = 6.34 m [S] Rx = - 4.99 m [E] [W] + 13.6 m [W] 8.61 m [W] = 8.6 m [W] Ry = 9.80 m [N] +- 6.34 m [S] [N] 3.46 m [N] Px Py 27o P Q Qy -Qx 25o 25o -Qy Qx -Q Because the directions for -Qx & -Qy have flipped, the resultant components change too!

R = P + (-Q) R2 = Rx2 + Ry2 = (8.6)2 + (3.46)2 = 73.96 + 11.9716 = (8.6)2 + (3.46)2 = 73.96 + 11.9716 R = √85.9316 = 9.3 m [W Ѳ N] Ѳ = tan-1 Ry = tan-1 3.46 = 22o Rx 8.6 R = 9.3 m [W22oN] X Y Px =(sin Ѳ) (hyp) = (sin 27o) (11.0 m) = 4.99 m [E] Py =(cos Ѳ) (hyp) = (cos 27o) (11.0 m) = 9.80 m [N] -Qx =(cos Ѳ) (hyp) = (cos 25o) (15.0 m) = 13.6 m [W] -Qy =(sin Ѳ) (hyp) = (sin 25o) (15.0 m) = 6.34 m [S] Rx = - 4.99 m [E] [W] + 13.6 m [W] 8.61 m [W] = 8.6 m [W] Ry = 9.80 m [N] +- 6.34 m [S] [N] 3.46 m [N] Ry R Rx