Problem 101 - Precalculus

Question

Prove that \(\mathbf{u}+3(2 \mathbf{v}-\mathbf{u})=6 \mathbf{v}-2 \mathbf{u},\) showing carefully how all relevant properties and definitions enter the proof.

Step-by-Step Solution

Answer

Both sides of the equation are equal: \(\mathbf{u} + 3(2\mathbf{v} - \mathbf{u}) = 6\mathbf{v} - 2\mathbf{u}\).

Step 1: Distribute the Scalar

We start by distributing the scalar 3 over the vector expression within the parentheses. The expression is \(3(2\mathbf{v} - \mathbf{u})\). Apply the distributive property: \(3 \times 2\mathbf{v} - 3 \times \mathbf{u} = 6\mathbf{v} - 3\mathbf{u}\). The expression becomes \(\mathbf{u} + 6\mathbf{v} - 3\mathbf{u}\).

Step 2: Combine Like Terms

Now, combine the like terms in the expression \(\mathbf{u} + 6\mathbf{v} - 3\mathbf{u}\). The like terms \(\mathbf{u}\) and \(-3\mathbf{u}\) can be combined as \(-2\mathbf{u}\) (i.e., \(1\mathbf{u} - 3\mathbf{u}\)). Thus, the expression simplifies to \(6\mathbf{v} - 2\mathbf{u}\).

Step 3: Compare with the Right-Hand Side

The simplified expression from Step 2 is \(6\mathbf{v} - 2\mathbf{u}\). This matches exactly with the expression given on the right-hand side of the original equation: \(6\mathbf{v} - 2\mathbf{u}\). Thus, we have shown that both sides of the equation are equal.

Key Concepts

Distributive Property

The distributive property is a fundamental rule in algebra that simplifies expressions and allows us to break down complex problems. When applied to vectors, it makes handling expressions like scalar multiplications straightforward. For example, given a scenario where you have a scalar

• Multiply the scalar by each term inside a vector expression
• Add the results together

This is exactly what we see in the solution of the given exercise. Initially, you have an expression within the parentheses, which looks like this: \(3(2\mathbf{v} - \mathbf{u})\).
By applying the distributive property, we distribute the scalar 3 across each term:

• First term: \(3 \times 2\mathbf{v} = 6\mathbf{v}\)
• Second term: \(3 \times (-\mathbf{u}) = -3\mathbf{u}\)

Hence, the expression simplifies to \(6\mathbf{v} - 3\mathbf{u}\).
This step is crucial because it breaks down the problem into manageable parts, simplifying what could otherwise be a complicated process.

Combining Like Terms

Combining like terms is an essential skill in simplifying algebraic and vector expressions. It involves merging terms that have the same variables and coefficients.
When dealing with vectors, terms like \(\mathbf{u}\) and \(-3\mathbf{u}\) are considered like terms because they involve the same vector \(\mathbf{u}\).

• Start by identifying similar terms in your expression.
• Add or subtract these terms together.

In our exercise, after applying the distributive property, we are left with the expression \(\mathbf{u} + 6\mathbf{v} - 3\mathbf{u}\).
Here, \(\mathbf{u}\) and \(-3\mathbf{u}\) are like terms. So, they combine to give \(-2\mathbf{u}\), calculated as follows:

• \(1\mathbf{u} - 3\mathbf{u} = -2\mathbf{u}\)

Thus, the expression now becomes \(6\mathbf{v} - 2\mathbf{u}\).
This process highlights the importance of careful calculation and organization when handling vectors.

Scalar Multiplication

Scalar multiplication refers to the multiplication of a vector by a scalar (a real number). This operation scales the vector by the scalar value, altering its magnitude without changing its direction.
In vector expressions, each component of the vector is multiplied by the scalar. With the exercise provided, the scalar is the number 3, which affects the vector expression \(2\mathbf{v} - \mathbf{u}\).

• Multiply each component of the vector by the scalar.
• For \(2\mathbf{v}\), it becomes \(3 \times 2\mathbf{v} = 6\mathbf{v}\).
• For \(-\mathbf{u}\), it becomes \(3 \times (-\mathbf{u}) = -3\mathbf{u}\).

This gives us a new vector expression \(6\mathbf{v} - 3\mathbf{u}\).
Through scalar multiplication, we adjust the magnitude of each vector component, which was pivotal in simplifying the original problem.