Chapter 7: Problem 20
If \(a=\left\langle a_{1}, a_{2}\right\rangle, b=\left\langle b_{1}, b_{2}\right\rangle, c=\left\langle c_{1}, c_{2}\right\rangle,\) and \(m\) and \(n\) are real numbers, prove the stated property. $$(m+n) \mathbf{a}=m \mathbf{a}+n \mathbf{a}$$
Short Answer
Expert verified
The property \((m+n) \mathbf{a} = m \mathbf{a} + n \mathbf{a}\) holds true.
Step by step solution
01
Understanding Vector and Scalar Multiplication
We begin with the expression \((m+n) \mathbf{a}\). This represents the multiplication of a scalar \((m+n)\) with the vector \(\mathbf{a} = \langle a_1, a_2 \rangle\). When a vector is multiplied by a scalar, each component of the vector is multiplied by that scalar.
02
Applying Scalar Multiplication to Vector \(a\)
Applying the scalar \((m+n)\) to the vector \(\mathbf{a}\), we get \((m+n)\mathbf{a} = \langle (m+n)a_1, (m+n)a_2 \rangle\). This expands both components of the vector \(\mathbf{a}\) by the scalar \((m+n)\).
03
Express \(m \mathbf{a} + n \mathbf{a}\) Separately
Now consider the expressions \(m \mathbf{a}\) and \(n \mathbf{a}\) separately. For \(m \mathbf{a}\), we have \(m \mathbf{a} = \langle ma_1, ma_2 \rangle\). Similarly, for \(n \mathbf{a}\), we have \(n \mathbf{a} = \langle na_1, na_2 \rangle\).
04
Combine Scalar Multiplied Vectors
Adding the vectors obtained in Step 3, we find: \(m \mathbf{a} + n \mathbf{a} = \langle ma_1, ma_2 \rangle + \langle na_1, na_2 \rangle\). Using vector addition, this results in \(\langle (ma_1 + na_1), (ma_2 + na_2) \rangle\) = \(\langle (m+n)a_1, (m+n)a_2 \rangle\).
05
Conclude the Equality
Comparing the results from Step 2 and Step 4, we observe that \((m+n) \mathbf{a} = \langle (m+n)a_1, (m+n)a_2 \rangle\) is equal to \(m \mathbf{a} + n \mathbf{a} = \langle (m+n)a_1, (m+n)a_2 \rangle\). Therefore, the property \((m+n) \mathbf{a} = m \mathbf{a} + n \mathbf{a}\) is proven to be true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Scalar Multiplication
Scalar multiplication is a fundamental operation in vector algebra that involves multiplying a vector by a scalar. A scalar is simply a real number, such as 3, -2, or 0.5. When a vector is multiplied by a scalar, each component of the vector is scaled by that number. Imagine stretching or compressing the vector in space.
For example, if you have a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), multiplying it by a scalar \( m \) results in a new vector \( m\mathbf{a} = \langle ma_1, ma_2 \rangle \). Here each component is multiplied by \( m \). This operation is consistent and applies for any number of dimensions.
For example, if you have a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), multiplying it by a scalar \( m \) results in a new vector \( m\mathbf{a} = \langle ma_1, ma_2 \rangle \). Here each component is multiplied by \( m \). This operation is consistent and applies for any number of dimensions.
- This means if \( m > 1 \), the vector gets stretched.
- If \( 0 < m < 1 \), the vector gets compressed.
- If \( m < 0 \), the vector direction reverses.
Distributive Property
The distributive property in vector algebra demonstrates how scalars distribute across vector addition. It’s a property from basic arithmetic that finds its parallel in vector math. This property can be represented as \( m(\mathbf{a} + \mathbf{b}) = m\mathbf{a} + m\mathbf{b} \). Now, let's connect this with our specific context.
For the property \((m+n) \mathbf{a} = m \mathbf{a} + n \mathbf{a}\), the distributive property is highlighted beautifully. Begin by considering a sum of two scalars \( (m+n) \) which multiplies a vector \( \mathbf{a} \). Applying this scalar multiplies each component of the vector. Now, separate the distribution into two parts: \( m\mathbf{a} \) and \( n\mathbf{a} \).
Here’s a breakdown:
For the property \((m+n) \mathbf{a} = m \mathbf{a} + n \mathbf{a}\), the distributive property is highlighted beautifully. Begin by considering a sum of two scalars \( (m+n) \) which multiplies a vector \( \mathbf{a} \). Applying this scalar multiplies each component of the vector. Now, separate the distribution into two parts: \( m\mathbf{a} \) and \( n\mathbf{a} \).
Here’s a breakdown:
- Compute \( m\mathbf{a} = \langle ma_1, ma_2 \rangle \).
- Compute \( n\mathbf{a} = \langle na_1, na_2 \rangle \).
- Combine these by addition: \( m\mathbf{a} + n\mathbf{a} = \langle ma_1 + na_1, ma_2 + na_2 \rangle \).
Vector Components
Vectors are entities defined by their components, which specify direction and magnitude. A 2-dimensional vector is often written as \( \mathbf{a} = \langle a_1, a_2 \rangle \), where \( a_1 \) and \( a_2 \) describe the vector’s position along the x and y axes, respectively. These components are critical when performing operations like addition, subtraction, and scalar multiplication.
In our exercise, understanding vector components is essential in transforming abstract scalar multiplication into a tangible calculation. Each component \( a_1 \) and \( a_2 \) in the vector \( \mathbf{a} \) is multiplied separately by the scalars \( m \) and \( n \).
In our exercise, understanding vector components is essential in transforming abstract scalar multiplication into a tangible calculation. Each component \( a_1 \) and \( a_2 \) in the vector \( \mathbf{a} \) is multiplied separately by the scalars \( m \) and \( n \).
- This means: \( m\mathbf{a} = \langle ma_1, ma_2 \rangle \) and \( n\mathbf{a} = \langle na_1, na_2 \rangle \).
- The result of vector addition \( m\mathbf{a} + n\mathbf{a} \) combines the respective components: \( \langle ma_1 + na_1, ma_2 + na_2 \rangle \).