For any two numbers a and b, the product of a−b times itself is equal to a2−2ab+b2.
Does this familiar algebraic result hold for dot products of a vector u − v with itself? In other words, is it true that
(u − v) • (u − v) = u•u−2u•v+v•v? Justify your conclusion,
trying not to express vectors u and v in component form.
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Use the property
a dot (b+c) = a dot b + a dot c
Yes, because of the dot product
To determine whether the algebraic result holds true for dot products of vectors, we need to apply the properties and definition of dot product without explicitly expressing vectors u and v in component form.
The dot product of two vectors u and v is defined as the sum of the products of their corresponding components:
u • v = u₁v₁ + u₂v₂ + u₃v₃ + ... + uₙvₙ.
Now let's examine the expression (u - v) • (u - v). Expanding it using the distributive property, we get:
(u - v) • (u - v) = (u - v)₁(u - v)₁ + (u - v)₂(u - v)₂ + (u - v)₃(u - v)₃ + ... + (u - v)ₙ(u - v)ₙ.
Using the distributive property again, we can expand each term:
(u - v) • (u - v) = (u₁ - v₁)² + (u₂ - v₂)² + (u₃ - v₃)² + ... + (uₙ - vₙ)².
Now, let's compare this result with the expression u•u - 2u•v + v•v. We can directly apply the definition of the dot product to this expression:
u•u - 2u•v + v•v = u₁u₁ + u₂u₂ + u₃u₃ + ... + uₙuₙ - 2(u₁v₁ + u₂v₂ + u₃v₃ + ... + uₙvₙ) + v₁v₁ + v₂v₂ + v₃v₃ + ... + vₙvₙ.
By rearranging the terms in the expanded expression, we can see that:
u•u - 2u•v + v•v = (u₁u₁ + u₂u₂ + u₃u₃ + ... + uₙuₙ) - 2(u₁v₁ + u₂v₂ + u₃v₃ + ... + uₙvₙ) + (v₁v₁ + v₂v₂ + v₃v₃ + ... + vₙvₙ).
Comparing this with the expanded form of (u - v) • (u - v), we can see that they are equal:
(u - v) • (u - v) = u•u - 2u•v + v•v.
Hence, we can conclude that the familiar algebraic result holds true for dot products of vectors, regardless of whether we express the vectors in component form or not.