Problem 17

Question

Pick any vectors $\vec{u}, \vec{v}$ and $\vec{w}$ in $\mathbb{R}^{3}$ and show that $\vec{u} \times(\vec{v}+\vec{w})=$ $\vec{u} \times \vec{v}+\vec{u} \times \vec{w}$

Step-by-Step Solution

Verified

Answer

The vector cross product distributes: $ \vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w} $.

1Step 1: Define the Vectors

Let's define $ \vec{u} = \begin{pmatrix} u_1 \ u_2 \ u_3 \end{pmatrix}, \vec{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}, $ and $ \vec{w} = \begin{pmatrix} w_1 \ w_2 \ w_3 \end{pmatrix} $. These are three-dimensional vectors with components in $ \mathbb{R}^3 $.

2Step 2: Express Vector Addition

Calculate $ \vec{v} + \vec{w} $, which gives: $ \vec{v} + \vec{w} = \begin{pmatrix} v_1 + w_1 \ v_2 + w_2 \ v_3 + w_3 \end{pmatrix} $.

3Step 3: Calculate Cross Product $ \vec{u} \times (\vec{v} + \vec{w}) $

The cross product $ \vec{u} \times (\vec{v} + \vec{w}) $ is given by the determinant of the matrix:\[ \small \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ u_1 & u_2 & u_3 \ v_1 + w_1 & v_2 + w_2 & v_3 + w_3 \end{vmatrix} \]. Calculating this determinant, we find: $ \begin{pmatrix} u_2(v_3 + w_3) - u_3(v_2 + w_2) \ u_3(v_1 + w_1) - u_1(v_3 + w_3) \ u_1(v_2 + w_2) - u_2(v_1 + w_1) \end{pmatrix} $.

4Step 4: Calculate Cross Products Separately

Calculate $ \vec{u} \times \vec{v} $:\[ \small \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} = \begin{pmatrix} u_2v_3 - u_3v_2 \ u_3v_1 - u_1v_3 \ u_1v_2 - u_2v_1 \end{pmatrix} \]. Calculate $ \vec{u} \times \vec{w} $:\[ \small \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ u_1 & u_2 & u_3 \ w_1 & w_2 & w_3 \end{vmatrix} = \begin{pmatrix} u_2w_3 - u_3w_2 \ u_3w_1 - u_1w_3 \ u_1w_2 - u_2w_1 \end{pmatrix} \].

5Step 5: Add Cross Products Together

Add the results from Step 4: $ (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1) $+ $ (u_2w_3 - u_3w_2, u_3w_1 - u_1w_3, u_1w_2 - u_2w_1) $.The resulting vector is: $ (u_2(v_3 + w_3) - u_3(v_2 + w_2), u_3(v_1 + w_1) - u_1(v_3 + w_3), u_1(v_2 + w_2) - u_2(v_1 + w_1)) $.

6Step 6: Compare Results

The expression from Step 5 matches the result of Step 3. Therefore, $ \vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w} $, confirming the vector cross product distributive property.

Key Concepts

Vectors in $\mathbb{R}^3$Vector AdditionDeterminantDistributive Property

Vectors in $\mathbb{R}^3$

In three-dimensional space, vectors are entities described by their three components in the coordinate form $ \begin{pmatrix} x \ y \ z \end{pmatrix} $. These components correspond to the x, y, and z axes respectively.
Vectors in $\mathbb{R}^3$ allow us to perform various operations such as addition and cross products, which have geometric and algebraic significance. The dimension "3" refers to the fact that they have three components designed to model real-world phenomena like velocity and force.
Understanding vectors in this space is crucial for calculating phenomena that exist in our 3-d world. By defining vectors, as in the exercise, $ \vec{u}, \vec{v}, $ and $ \vec{w} $, we can proceed to perform operations that demonstrate fundamental mathematical properties.

Vector Addition

Vector addition is the process of adding two vectors by adding their respective components. For instance, if we have two vectors $ \vec{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix} $ and $ \vec{w} = \begin{pmatrix} w_1 \ w_2 \ w_3 \end{pmatrix} $, the sum $ \vec{v} + \vec{w} $ is given by combining each of the respective components:
$ \vec{v} + \vec{w} = \begin{pmatrix} v_1 + w_1 \ v_2 + w_2 \ v_3 + w_3 \end{pmatrix} $.
This technique of addition is intuitive as it visually represents how, in three-dimensional space, the resulting vector combines the influences (like direction and magnitude) of the individual vectors.

Determinant

The determinant is a special number that can be calculated from the elements of a square matrix. In the context of vectors, it helps in computing the cross product. The cross product $ \vec{u} \times \vec{v} $ can be determined by evaluating a 3x3 matrix determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \],
where $ \hat{i}, \hat{j}, \hat{k} $ are unit vectors along the x, y, and z axes, respectively.
The determinant expansion along the first row provides a systematic way to calculate the result, turning components into a new vector that is perpendicular to the original vectors, an essential feature in three-dimensional analyses.

Distributive Property

The distributive property in vector algebra particularly emphasizes how operations distribute across sums. Specifically, for cross products, this property states that:
$ \vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w} $.
To understand this, consider that the cross product itself is a binary operation producing a vector perpendicular to both input vectors. The distribution of $ \vec{u} $ across the addition $ \vec{v} + \vec{w} $ maintains this uniqueness.
Proving this property involves computing the left-hand side and the right-hand side independently and showing their equivalence, as shown in the exercise. This property is vital because it allows simplifications in vector analysis, crucial in physics and engineering.

Problem 17