Convert the following vectors to Cartesian form; (2 marks each)
1. 13 m on a bearing of 165°
2. 120 km/h 20° West of North
1. X
Y
Using soh cah toa
X= 13cos(165-90)
X= 3.3646475863
X= 3.4m
Y= 13sin(165-90)
Y= 12.5570357419
Y= 12.6m
therefore, cartesian vector is [3.4,-12.6]
2. X
Y
X= 120sin20
X= 41.04241716
X= 41km/h
Y= 120cos20
Y= 112.763114496
Y= 113km\h Therefore, cartesian vector is [41,-113]
Convert the following vectors to direction/magnitude form, giving the direction as a bearing; (2 marks each)
1. [12, 5]
2. [-31, -11]
1.
A
A= square root (12^2 + 5^2)
A= 13 units
Tan-1 (5/12) = 22.619864948°
90 - 22.619864948 = 67.380135052°
Therefore, 13 units a bearing of 67°
2.
B
B = square root ((-31)^2+(-11)^2)
B = 32.8937684068
Tan-1 (-11/-31) = 19.5366549381
Therefore, 33 units a bearing of 251°