If #a^2 + b^2 =1# and #m^2+n^2=1#, then which of the following is true for all values of #m,n,a,b#?(the question has multiple answers)

A)#|am+bn|<=1#
B)#|am-bn|>=1#
C)#|am-bn|<=1#

1 Answer
Oct 26, 2017

# abs(am+bn) le 1 #

Explanation:

Calling

#vec u = (a,b)#
#vec v = (m,n)#

such that

#norm(vec u) = sqrt(a^2+b^2) = 1#
#norm(vec v) = sqrt(m^2+n^2) = 1#

we have

# << vec u, vec v >> =(am+bn)= norm(vec u) norm(vec v) cos (hat(vec u, vec v)) =cos (hat(vec u, vec v))#

and as we know #-1 le cos (hat(vec u, vec v)) le 1# or

#abs(cos (hat(vec u, vec v))) le 1 hArr abs(am+bn) le 1 #