2006-09-23 05:07:29 · 4 answers · asked by Jayu P 1 in Science & Mathematics ➔ Physics
|A+B| = sqrt(2*A*A + 2*A*Acos(i))
|A-B| = sqrt(2*A*A-2*A*Acos(i))
so, (1+cos(i)) = 110*110 (1-cos(i))
or cos(i) = (110^2-1)/(110^2+1) = 0.999
or i = 1.04 degrees
2006-09-23 06:02:53 · answer #1 · answered by ag_iitkgp 7 · 1⤊ 0⤋
Since the magnitude of vector(A-B) =magnitude of vector (A-B)
A^2 +B^2 + 2A.B =A^2 +B^2 - 2A.B
therefore 4A.B=0 which means are perpendicular to each other or angle between them is 90 degree.
2006-09-23 13:05:22 · answer #2 · answered by Amar Soni 7 · 0⤊ 0⤋
I do not think that these vectors could have equal magnitudes.
2006-09-23 12:19:39 · answer #3 · answered by bruinfan 7 · 0⤊ 0⤋
0.0181813 radians or 3.123 degrees. At this angle, cos(theta/2)/sin(theta/2)=110. With unit-length vectors, A+B=2cos(theta/2) and A-B=2sin(theta/2).
2006-09-23 12:59:35 · answer #4 · answered by kirchwey 7 · 0⤊ 0⤋