Find the angle between the vectors u and v if u = (-1,4) v = (3,-4)
a. 155.93
b. 158.58
c. 158.13
d. 155.24
e. 157.17
+118659 u = (-1,4) v = (3,-4)
I am sure there is a less clumsy way to do this but....
u is in the second quadrant. The angle it makes with the positive x axis is
\(\pi-atan(4/1) = \pi-atan(4)\)
v is in the 4th quadrant so the angle it makes with the positive x axis is
\(atan(-4/3)\)
\(angle\;\;between=|180-atan(4)-atan(-4/3)|\\\)
104.036243467926-atan(-4/3) = 157.166345822082
157.17 degrees
+118659 u = (-1,4) v = (3,-4)
I am sure there is a less clumsy way to do this but....
u is in the second quadrant. The angle it makes with the positive x axis is
\(\pi-atan(4/1) = \pi-atan(4)\)
v is in the 4th quadrant so the angle it makes with the positive x axis is
\(atan(-4/3)\)
\(angle\;\;between=|180-atan(4)-atan(-4/3)|\\\)
104.036243467926-atan(-4/3) = 157.166345822082
157.17 degrees
+129841 cos (theta) = [ u (dot) v] / [ ll u ll * ll v ll ]
u (dot) v = -1*3 + 4 * -4 = -19
ll u ll = sqrt [ (-1)^2 + 4^2 ] = sqrt (17)
ll v ll = sqrt [ (3)^2 + (-4)^2 ] = 5
So
cos (theta) = -19 / [ 5 sqrt (17)] so using the cosine inverse to find theta, we have
arccos ( -19 / [ 5 sqrt (17)] ) = theta = about 157.166° ...... "e" is correct
