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Chapter 6: Problem 17

Show that the scalar product obeys the distributive law: \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\)

Short Answer

Expert verified
In conclusion, you have shown that the scalar product \( \vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} \), thus the scalar product obeys the distributive law. This claim is based on how scalar products operate and how vectors work.

Step by step solution

01

Understand Scalar Product of Vectors

Firstly, define the scalar product. It is also called the dot product. It's an operation that combines two vectors to yield a scalar (or number) instead of another vector. It's symbolized by a dot '·' between two vectors.
02

Use the Definition of Scalar Product on Vectors with Components

Suppose, the vectors \( \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \), \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \), and \( \vec{C}= C_x \hat{i} + C_y \hat{j} + C_z \hat{k} \). Using the distributive law, calculate the scalar product of \( \vec{A} \) with \( \vec{B}+\vec{C} \) as follows: \( \vec{A} · (\vec{B} + \vec{C}) = \vec{A} · \vec{B} + \vec{A} · \vec{C} \)
03

Apply this expression to the respective vectors

Now, apply the components of vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) into the distributive law formula mentioned in step 2. As such, the dot product of two vectors can be expressed as the product of the magnitudes of the two vectors and the cosine of the angle between them.
04

Finalize your proof

Showing that the left-hand side equals the right-hand side for all vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) - by comparing corresponding components on both sides if vectors are expressed in Cartesian coordinates - you can confirm that the scalar product indeed follows the distributive law. If you want to take it one step further, you might want to provide examples of vectors and calculate the scalar product in both ways to demonstrate that the results are indeed identical.

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