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Chapter 6: Problem 32

Show that if \(\mathbf{u}\) and \(\mathbf{v}\) are vectors, then $$ |\mathbf{u}+\mathbf{v}|^{2}=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2} $$.

Short Answer

Expert verified
To prove the given identity, begin by expanding the magnitude expression: \(|\mathbf{u}+\mathbf{v}|^{2} = (\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}+\mathbf{v})\). Distribute the dot product and simplify the terms, using properties of the dot product and magnitude: \(|\mathbf{u}|^2 + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{u} + |\mathbf{v}|^2\). Using the commutative property of the dot product, rewrite the second and third terms: \(|\mathbf{u}|^2 + \mathbf{u}\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^2\). Finally, combine the two middle terms to complete the proof: \(|\mathbf{u}+\mathbf{v}|^{2} = |\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\).

Step by step solution

01

Expand the magnitude expression

Begin by expanding the magnitude expression on the left side of the equation: $$ |\mathbf{u}+\mathbf{v}|^{2}= (\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}+\mathbf{v}) $$ Here, we have used the fact that the square of the magnitude of a vector is equal to the dot product of the vector with itself.
02

Distribute the dot product

Now, distribute the dot product using the properties of the dot product: $$ (\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}+\mathbf{v})= \mathbf{u}\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{u} + \mathbf{v}\cdot\mathbf{v} $$ We have used the distributive property of the dot product.
03

Simplify the terms

Simplify the terms obtained in the previous step using the properties of the dot product and magnitude: $$ \mathbf{u}\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{u} + \mathbf{v}\cdot\mathbf{v} = |\mathbf{u}|^2 + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{u} + |\mathbf{v}|^2 $$ Here, we have used the fact that \(\mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2\) and \(\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2\).
04

Use commutative property

Since the dot product is commutative, we can rewrite the second and third terms as follows: $$ |\mathbf{u}|^2 + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{u} + |\mathbf{v}|^2 = |\mathbf{u}|^2 + \mathbf{u}\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^2 $$ This is true because \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\).
05

Combine terms and complete the proof

Now combine the two middle terms and we have the final expression: $$ |\mathbf{u}|^2 + \mathbf{u}\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^2 = |\mathbf{u}|^2 + 2 \mathbf{u} \cdot \mathbf{v} + |\mathbf{v}|^2 $$ Hence, we have proven that $$ |\mathbf{u}+\mathbf{v}|^{2}=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2} $$

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