Recent content by ios

don't i need to maintain that x to y ratio for the angle of vector c to be perpendicular to vector a? i thought that was the only way to keep the angle so that added together, the angle between vector c and vector a equals 90.

i'm only given s1, theta1, s2, theta2, s3, theta3 :rolleyes:

don't i need that for vector c to be perpendicular to vector a?

A cross-country skier skis s1 in the direction theta1 west of south, then s2 in the direction theta2 north of east, and finally s3 in the direction theta3 south of west. How far is the skier from the starting point? I drew it..and it looks really really weird, I haven't done physics in a...

Xc is the x component of vector c, and Yc is the y component of vector c. I'm just not sure if my math was right or not. 32.5372 is the angle if you take tan of 4.5/7. and 1.556 is the ratio of x to y (vector a).

You are given vectors a = 4.50i-7.00j and b = -3.50i+7.40j. A third vector c lies in the xy-plane. Vector c is perpendicular to vector a and the scalar product of c with b is 10.0. now i figured that since vector a is in quad 4, then to be perpendicular, vector c would have to be in quad 1...

oh sorry, didnt see the homework part. thanks thanks

i'm stumped totally on this physics question. help please? vectors a and b are drawn from the same origin with an angle seperating the two vectors. if a and b have the same magnitude, what value of an angle will their vector sum have the same magnitude as vectors a or b? i've tried over...