Question 1044123: Is the following statement always, sometimes or never true? Justify your reasoning
there should be arrows above the letters
|u + v| > |u - v|
thanks
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Let's set up some notation
a,b,c,d are scalars
u and v are vectors such that
u = (a,b)
v = (c,d)
So that means
u+v = (a+c,b+d)
u-v = (a-c,b-d)
Now use the length of vector formula to get
|u + v| = sqrt( (a+c)^2 + (b+d)^2 )
|u + v| = sqrt( a^2+2ac+c^2 + b^2+2bd+d^2 )
and this as well
|u - v| = sqrt( (a-c)^2 + (b-d)^2 )
|u - v| = sqrt( a^2-2ac+c^2 + b^2-2bd+d^2 )
Now let's go back to
|u + v| > |u - v|
Perform substitutions and then square both sides to get
|u + v| > |u - v|
sqrt( a^2+2ac+c^2 + b^2+2bd+d^2 ) > sqrt( a^2-2ac+c^2 + b^2-2bd+d^2 )
a^2+2ac+c^2 + b^2+2bd+d^2 > a^2-2ac+c^2 + b^2-2bd+d^2
Things look messy, but we can subtract a^2, c^2, b^2 and d^2 from both sides to have those terms cancel out. We will be left with this
2ac + 2bd > -2ac - 2bd
add 2ac and 2bd to both sides and you'll get
2ac+2bd+2ac+2bd > 0
4ac+4bd > 0
which ultimately simplifies to ac+bd > 0
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So if |u+v| > |u-v|, where u and v are defined above, then ac+bd > 0. The same can be said in reverse as well.
There are some cases where ac+bd > 0 is true. But it could be false as well.
For instance, if a = 2, b = 3, c = 1 and d = 2, then the inequality would be true.
Compare this to if a = 2, b = 3, c = -1 and d = -2, then the inequality would be false.
Side note: the inequality is only true if the angle between the vectors u and v is acute. If the angle is obtuse, then the inequality is false.
So the final answer is sometimes
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