Use the vector form of the Law of Cosines to show that |u + v|^2 ≤ (|u| + |v|)^2 holds for
any vectors u and v. What does this prove?
To show that |u + v|^2 ≤ (|u| + |v|)^2 holds for any vectors u and v, we can use the vector form of the Law of Cosines.
The Law of Cosines states that for any triangle with sides of length |u|, |v|, and |u + v|, the following equation holds:
|u + v|^2 = |u|^2 + |v|^2 - 2|u||v|cosθ
where θ is the angle between vectors u and v.
Now, let's substitute the equation of the Law of Cosines into the inequality we need to prove:
|u + v|^2 ≤ (|u| + |v|)^2
Substituting |u + v|^2 = |u|^2 + |v|^2 - 2|u||v|cosθ, the inequality becomes:
|u|^2 + |v|^2 - 2|u||v|cosθ ≤ (|u| + |v|)^2
Expanding the right-hand side of the inequality, we have:
|u|^2 + |v|^2 - 2|u||v|cosθ ≤ |u|^2 + 2|u||v| + |v|^2
By cancelling out the common terms on both sides, the inequality simplifies to:
-2|u||v|cosθ ≤ 2|u||v|
Dividing both sides of the inequality by 2|u||v| (since |u||v| ≠ 0), we get:
- cosθ ≤ 1
Since the range of values for the cosine function is -1 ≤ cosθ ≤ 1, the inequality holds true.
Therefore, |u + v|^2 ≤ (|u| + |v|)^2 is proved.
This inequality shows that the magnitude squared of the sum of two vectors is always less than or equal to the squared sum of their magnitudes.
To prove that |u + v|^2 ≤ (|u| + |v|)^2 using the vector form of the Law of Cosines, we start by expressing |u + v|^2 in terms of u and v:
|u + v|^2 = (u + v) · (u + v)
By the distributive property of dot product, we can expand this expression:
= u · u + u · v + v · u + v · v
Now, according to the Law of Cosines, the dot product of two vectors can be expressed in terms of their magnitudes and the cosine of the angle between them:
u · v = |u| |v| cos(theta)
Substituting this into our expression, we get:
= |u|^2 + 2 u · v + |v|^2
Now, let's focus on the right side of the inequality: (|u| + |v|)^2. Expanding this expression, we have:
(|u| + |v|)^2 = |u|^2 + 2|u||v| + |v|^2
Comparing the two expressions, we see that they both have the same terms: |u|^2, 2 u · v, and |v|^2.
Since cos(theta) can take values between -1 and 1, we know that |u|^2 + 2 u · v + |v|^2 is less than or equal to |u|^2 + 2|u||v| + |v|^2.
Thus, we have shown that |u + v|^2 ≤ (|u| + |v|)^2 for any vectors u and v.
This inequality proves something important. It shows that the magnitude of the sum of two vectors (u + v) squared is always less than or equal to the square of the sum of their magnitudes. In other words, the sum of the lengths (magnitudes) of two vectors is always greater than or equal to the length of their sum.
Geometrically, this inequality implies that the triangle formed by the vectors u, v, and u + v is always acute or at most right-angled. The inequality can be interpreted as saying that the sum of the squares of the sides of a triangle is always less than or equal to the square of the sum of the lengths of those sides.
Understanding and applying this inequality can be useful in various fields like geometry, physics, computer graphics, and many other areas where vectors are used.
Actually, it's even better than that:
|u + v|^2 ≤ |u|^2 + |v|^2 ≤ (|u|+|v|)^2