Compute the angle, in radians, between the two vectors:
A=5i - 8j
B=7i + 3j
2007-12-19 02:31:06 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Vector a = 5i - 8j
|a| = (5² + 8²)^(1/2) = 89^(1/2)
Vector b = 7i + 3j
|b| = (7² + 3²)^(1/2) = 58^(1/2)
a.b = (5i - 8j) . (7i + 3j) = 35 - 24 = 11
cos Ө = 11 / 89^(1/2) 58^(1/2)
Ө = 1.42 radians
2007-12-19 03:10:45 · answer #1 · answered by Como 7 · 2⤊ 1⤋
Formula :-
cosx = A dot B / ( | A |x| B | )
cosx=(5x7)+(-8x3) / â5²+8² x â7²+8²
x=82.209894555°
Angle = 82.209894555°
2007-12-19 10:35:10 · answer #2 · answered by Murtaza 6 · 0⤊ 1⤋
Find A.B , the dot product which has magnitude |A.B| = |A||B| cos m
m = angle between vectors in radians
So m = arccos (|A.B|/|A||B|) ... find it.
Also you may use the cross product and use arcsin...
2007-12-19 10:43:27 · answer #3 · answered by Deep B 2 · 0⤊ 1⤋
Okay I think its tan(8/5)+tan(3/7) = 1.0122+.40489 = 1.417089 approximately.
2007-12-19 11:15:28 · answer #4 · answered by yljacktt 5 · 0⤊ 1⤋
use dot product
A.B=|A||B|cosang
|A|=sqrt(5^2+8^2)=9.43398
|B|=sqrt(49+9)=7.615773
A.B=35-24=11
cosang=11/(9.43398X7.615773)=.1531
ang=1.417 radian
2007-12-19 10:45:31 · answer #5 · answered by ncoe 2 · 0⤊ 1⤋
the relation is a.b/[A][B]=cos x ([A] IS MAGNITUDE OF A)
2007-12-19 10:40:01 · answer #6 · answered by soumyo 4 · 0⤊ 1⤋