Problem 55 - Contemporary Precalculus

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. $$r(\mathbf{u}+\mathbf{v})=r \mathbf{u}+r \mathbf{v}$$

Step-by-Step Solution

Answer

Question: Prove that for a scalar r and vectors u and v, r(u + v)\(= ru + rv\). Answer: We proved that \(r(\mathbf{u}+\mathbf{v})=r\mathbf{u}+r\mathbf{v}\) by calculating the left side (\(\langle r(a+c), r(b+d) \rangle\)) and the right side (\(\langle ra+rc, rb+rd \rangle\)) of the equation and showing that they are equal, which implies that the stated vector property holds.

Step 1: Calculate the left side of the equation

To calculate the left side of the equation \(r(\mathbf{u}+\mathbf{v})\), first add the vectors \(\mathbf{u}\) and \(\mathbf{v}\), and then multiply the resulting vector by the scalar \(r\). $$r(\mathbf{u}+\mathbf{v}) = r\left(\langle a, b\rangle + \langle c,d \rangle\right)$$ Add the corresponding components of the vectors \(\langle a, b\rangle\) and \(\langle c,d \rangle\): $$r(\mathbf{u}+\mathbf{v}) = r\langle a+c, b+d \rangle$$ Multiply the vector \(\langle a+c, b+d \rangle\) by the scalar \(r\): $$r(\mathbf{u}+\mathbf{v}) = \langle r(a+c), r(b+d) \rangle$$

Step 2: Calculate the right side of the equation

To calculate the right side of the equation \(r\mathbf{u}+r\mathbf{v}\), first multiply the vectors \(\mathbf{u}\) and \(\mathbf{v}\) with the scalar \(r\), then add the resulting vectors. $$r\mathbf{u} = r\langle a, b \rangle = \langle ra, rb \rangle$$ $$r\mathbf{v} = r\langle c, d \rangle = \langle rc, rd \rangle$$ Now add the two vectors \(\langle ra, rb \rangle\) and \(\langle rc, rd \rangle\): $$r\mathbf{u}+r\mathbf{v} = \langle ra, rb \rangle + \langle rc, rd \rangle = \langle ra+rc, rb+rd \rangle$$

Step 3: Compare the left and right sides of the equation

Now let's compare the left and right sides of the equation. From Step 1, we found that the left side of the equation is: $$r(\mathbf{u}+\mathbf{v}) = \langle r(a+c), r(b+d) \rangle$$ From Step 2, we found that the right side of the equation is: $$r\mathbf{u}+r\mathbf{v} = \langle ra+rc, rb+rd \rangle$$ It can be observed that the left and right sides of the equation are equal: $$\langle r(a+c), r(b+d) \rangle = \langle ra+rc, rb+rd \rangle$$ Therefore, the stated vector property holds: $$r(\mathbf{u}+\mathbf{v})=r \mathbf{u}+r \mathbf{v}$$

Key Concepts

Scalars

Scalars are numerical values, plain and simple. When you see a number that doesn't have any direction, like 2 or -5, you're looking at a scalar. Scalars can be positive, negative, or even zero. They are used to measure quantities like temperature, mass, and time—basically anything that doesn't require direction.
Scalars also play an important role in vector arithmetic. They are often used to multiply vectors, a process that can stretch or shrink the vector. In vector addition exercises like the one given in the original problem, the scalars allow us to weigh vectors without changing their direction. This property is crucial for many applications in physics and engineering.
So when you see a scalar in the context of vector operations, it's basically telling you how much you're going to scale that vector. It's like turning up or down the volume on a stereo, but instead of making music louder or softer, you're making a vector longer or shorter.

Vector Operations

Vector operations include a variety of mathematical manipulations you can perform on vectors, such as addition, subtraction, and scalar multiplication. In our exercise, vector addition and scalar multiplication are crucial for solving the problem.
When you add vectors, you are essentially combining them. The addition is done component-wise: for two vectors \( \mathbf{u} = \langle a, b \rangle \) and \( \mathbf{v} = \langle c, d \rangle \), the sum is \( \mathbf{u} + \mathbf{v} = \langle a+c, b+d \rangle \). This is just like adding two lists of numbers, going one by one.
Scalar multiplication involves multiplying each component of a vector by a scalar. For a vector \( \mathbf{u} = \langle a, b \rangle \) and scalar \( r \), multiplying the vector by the scalar gives \( r \mathbf{u} = \langle ra, rb \rangle \).

• **Addition** is like mixing ingredients in a recipe to see what new flavors emerge.
• **Scalar Multiplication** is like stretching a rubber band further without changing its direction.

These operations are the building blocks of more complex vector manipulations, making them fundamental for anyone studying physics, engineering, or any field that uses vectors.

Distributive Property

In mathematics, the distributive property is a way to simplify expressions and understand how multiplication interacts with addition. This concept applies perfectly even in the world of vectors.
The distributive property in the context of vector addition and scalar multiplication means that for scalar \( r \) and vectors \( \mathbf{u} \) and \( \mathbf{v} \), the property \( r(\mathbf{u} + \mathbf{v}) = r\mathbf{u} + r\mathbf{v} \) holds true. Let's break it down:

• On the left side, you first add vectors \( \mathbf{u} \) and \( \mathbf{v} \), then multiply the resulting vector by the scalar \( r \).
• On the right side, you first scale each vector \( \mathbf{u} \) and \( \mathbf{v} \) individually by the scalar \( r \), and then you add these scaled vectors together.

Both methods give the same result because multiplication by the scalar distributes over the addition of vectors. Visualize it as distributing the influence of the scalar over the components of the resulting vector.
Understanding this property not only boosts your prowess in solving vector operations but also provides important insights into solving algebraic expressions and equations. It's a versatile tool that belongs in every math student's toolkit.