What conditions must be satisfied by vectors u and v for this to be true?
|u + v| >|u - v|
To determine the conditions under which the inequality |u + v| > |u - v| holds true, we need to analyze the properties of vectors and their magnitudes.
Let's break down the absolute value expressions for vectors u + v and u - v:
|u + v| = sqrt((u1 + v1)^2 + (u2 + v2)^2 + ... + (un + vn)^2)
|u - v| = sqrt((u1 - v1)^2 + (u2 - v2)^2 + ... + (un - vn)^2)
Now, let's compare these expressions:
|u + v| > |u - v|
Squaring both sides of the inequality, we get:
(u1 + v1)^2 + (u2 + v2)^2 + ... + (un + vn)^2 > (u1 - v1)^2 + (u2 - v2)^2 + ... + (un - vn)^2
Expanding and simplifying:
u1^2 + 2u1v1 + v1^2 + u2^2 + 2u2v2 + v2^2 + ... + un^2 + 2unvn + vn^2 >
u1^2 - 2u1v1 + v1^2 + u2^2 - 2u2v2 + v2^2 + ... + un^2 - 2unvn + vn^2
The squared terms on both sides of the inequality cancel out, resulting in:
2u1v1 + 2u2v2 + ... + 2unvn > - 2u1v1 - 2u2v2 - ... - 2unvn
This implies that the sum of the products of corresponding components of vectors u and v on the left-hand side of the inequality must be greater than the negation of the sum of the products on the right-hand side.
Therefore, the conditions for the inequality |u + v| > |u - v| to hold true would be:
2u1v1 + 2u2v2 + ... + 2unvn > - 2u1v1 - 2u2v2 - ... - 2unvn
Simplifying further:
u1v1 + u2v2 + ... + unvn > 0
In other words, the dot product of vectors u and v must be greater than zero for |u + v| to be greater than |u - v|.