Problem 56
Question
Let \(\boldsymbol{u}=\langle a, b\rangle\) and \(\boldsymbol{v}=\langle c, d\rangle,\) and let \(r\) and s be scalars. Prove that the stated property holds by calculating the vector on each side of the equal sign. $$(r+s) \mathbf{v}=r \mathbf{v}+s \mathbf{v}$$
Step-by-Step Solution
Verified Answer
Question: Prove that the given property $$(r+s)\mathbf{v} = r\mathbf{v} + s\mathbf{v}$$ is true for any vector \(\mathbf{v}\) and any scalars \(r\) and \(s\).
Answer: After performing calculations for each side of the equation and comparing them, we found that the left side $(r+s)\mathbf{v} = \langle (r+s)c, (r+s)d \rangle$ is equal to the right side $r\mathbf{v} + s\mathbf{v} = \langle (r+s)c, (r+s)d \rangle$. Therefore, we can conclude that the given property $$(r+s)\mathbf{v} = r\mathbf{v} + s\mathbf{v}$$ is true for any vector \(\mathbf{v}\) and any scalars \(r\) and \(s\).
1Step 1: Set up the given property
First, we need to write down the given property that we are going to prove: $$(r+s)\mathbf{v} = r\mathbf{v} + s\mathbf{v}$$ where \(\mathbf{v} = \langle c, d \rangle\), and \(r\) and \(s\) are scalars.
2Step 2: Expand the left side of the equation
Now let's expand the left side of the equation, which involves multiplying the vector \(\mathbf{v}\) by the scalar \((r+s)\). This results in: $$(r+s)\mathbf{v} = (r+s)\langle c, d \rangle = \langle (r+s)c, (r+s)d \rangle$$
3Step 3: Expand the right side of the equation
Next, we expand the right side of the equation, which involves adding the product of the vector \(\mathbf{v}\) multiplied by the scalars \(r\) and \(s\): $$r\mathbf{v} + s\mathbf{v} = r\langle c, d \rangle + s\langle c, d \rangle = \langle rc, rd \rangle + \langle sc, sd \rangle$$
4Step 4: Simplify the right side of the equation
Now, let's simplify the right side of the equation by adding the two vectors component-wise: $$\langle rc, rd \rangle + \langle sc, sd \rangle = \langle rc+sc, rd+sd \rangle = \langle (r+s)c, (r+s)d \rangle$$
5Step 5: Compare the two sides of the equation
Finally, let's compare the two sides of the equation. The left side is $$(r+s)\mathbf{v} = \langle (r+s)c, (r+s)d \rangle$$ and the right side is $$r\mathbf{v} + s\mathbf{v} = \langle (r+s)c, (r+s)d \rangle$$ Since both sides are equal, we've proven that the given property $$(r+s)\mathbf{v} = r\mathbf{v} + s\mathbf{v}$$ is true.
Key Concepts
Vector AdditionProof by CalculationDistributive Property
Vector Addition
Vector addition is a fundamental concept in vector algebra, involving the combination of two or more vectors to form a resultant vector. Imagine you're taking two trips, one after the other. The total distance and direction you've traveled can be represented as a single journey—a good analogy for vector addition.
When we add two vectors, we do so component-wise: \ and \, their sum is a new vector having components that are the sums of the corresponding components of \ and \, just as seen in Step 4 of our solution where we added vectors \
When we add two vectors, we do so component-wise: \
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- \(\langle a_1, b_1 \rangle + \langle a_2, b_2 \rangle = \langle a_1 + a_2, b_1 + b_2 \rangle\) \
Proof by Calculation
Proof by calculation is a way of verifying mathematical properties or theorems by working through the operations step by step, showing that the initial statement is valid. It requires a logical progression from one equation to the next, much like building a wall brick by brick, ensuring that each step is based on the preceding one without any gaps or errors.
In our textbook exercise, the step by step solution exemplifies a proof by calculation. Starting with the initial property, we meticulously expanded both sides of the equation, simplified where necessary, and ended up with two identical expressions, which conclusively proved the original statement. It's like confirming that two recipes yield the same delicious cake by comparing their ingredients and baking steps. This kind of proof is particularly satisfying because it allows you to follow the logic from start to finish.
In our textbook exercise, the step by step solution exemplifies a proof by calculation. Starting with the initial property, we meticulously expanded both sides of the equation, simplified where necessary, and ended up with two identical expressions, which conclusively proved the original statement. It's like confirming that two recipes yield the same delicious cake by comparing their ingredients and baking steps. This kind of proof is particularly satisfying because it allows you to follow the logic from start to finish.
Distributive Property
The distributive property is a cornerstone of algebra that allows us to simplify expressions and solve equations efficiently. It tells us how multiplication acts over addition (or subtraction). Specifically, multiplying a number by a sum is the same as multiplying each addend by the number and then adding the products together: \
In the context of vector scalar multiplication, it means that a scalar multiplied by the sum of two vectors is equivalent to multiplying the scalar by each individual vector and then adding the results. Our exercise demonstrates this by showing that the scalar multiplied by the sum of components results in the same vector as when we distribute the scalar among the components individually. It's like sprinkling seasoning over a mix of vegetables both before and after mixing them—the outcome in taste should be uniformly the same.
- \
- \(a(b + c) = ab + ac\) \
In the context of vector scalar multiplication, it means that a scalar multiplied by the sum of two vectors is equivalent to multiplying the scalar by each individual vector and then adding the results. Our exercise demonstrates this by showing that the scalar multiplied by the sum of components results in the same vector as when we distribute the scalar among the components individually. It's like sprinkling seasoning over a mix of vegetables both before and after mixing them—the outcome in taste should be uniformly the same.
Other exercises in this chapter
Problem 55
Let \(\boldsymbol{u}=\langle a, b\rangle\) and \(\boldsymbol{v}=\langle c, d\rangle,\) and let \(r\) and s be scalars. Prove that the stated property holds by c
View solution Problem 55
In Exercises \(53-64,\) perform the indicated multiplication or division. Express your answer in both polar form \(r(\cos \theta+i \sin \theta)\) and rectangula
View solution Problem 56
In Exercises \(53-64,\) perform the indicated multiplication or division. Express your answer in both polar form \(r(\cos \theta+i \sin \theta)\) and rectangula
View solution Problem 57
Let \(\boldsymbol{u}=\langle a, b\rangle\) and \(\boldsymbol{v}=\langle c, d\rangle,\) and let \(r\) and s be scalars. Prove that the stated property holds by c
View solution