Problem 41 - Thomas Calculus

Question

Linear combination Let \(\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \mathbf{v}=\mathbf{i}+\mathbf{j},\) and \(\mathbf{w}=\) \(\mathbf{i}-\mathbf{j} .\) Find scalars \(a\) and \(b\) such that \(\mathbf{u}=a \mathbf{v}+b \mathbf{w}.\)

Step-by-Step Solution

Answer

The scalars are \(a = \frac{3}{2}\) and \(b = \frac{1}{2}\).

Step 1: Understanding the Problem

We need to express the vector \(\mathbf{u} = 2\mathbf{i} + \mathbf{j}\) as a linear combination of \(\mathbf{v} = \mathbf{i} + \mathbf{j}\) and \(\mathbf{w} = \mathbf{i} - \mathbf{j}\). This means finding scalars \(a\) and \(b\) such that \(\mathbf{u} = a\mathbf{v} + b\mathbf{w}\).

Step 2: Express the Linear Combination

Express the linear combination using the given vectors: \[a \mathbf{v} + b \mathbf{w} = a(\mathbf{i} + \mathbf{j}) + b(\mathbf{i} - \mathbf{j}).\] Simplifying, we combine like terms: \[a\mathbf{i} + a\mathbf{j} + b\mathbf{i} - b\mathbf{j} = (a + b)\mathbf{i} + (a - b)\mathbf{j}.\]

Step 3: Set Up the Equations

Equate the vector expression to \(\mathbf{u}\): \[(a + b)\mathbf{i} + (a - b)\mathbf{j} = 2\mathbf{i} + \mathbf{j}.\] This results in two equations: \(a + b = 2\) and \(a - b = 1\).

Step 4: Solve for Scalars

Solve the system of equations: \[\begin{align*} a + b &= 2, \ a - b &= 1. \end{align*}\] Add the two equations to eliminate \(b\): \[2a = 3 \Rightarrow a = \frac{3}{2}.\] Substitute \(a\) back into one of the original equations: \[a + b = 2 \Rightarrow \left(\frac{3}{2}\right) + b = 2 \Rightarrow b = \frac{1}{2}.\]

Step 5: Verify the Solution

Substitute the values of \(a\) and \(b\) back into the expression: \[a \mathbf{v} + b \mathbf{w} = \frac{3}{2}(\mathbf{i} + \mathbf{j}) + \frac{1}{2}(\mathbf{i} - \mathbf{j}) = \frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + \frac{1}{2}\mathbf{i} - \frac{1}{2}\mathbf{j}.\] Simplifying gives: \[2\mathbf{i} + \mathbf{j},\] which matches \(\mathbf{u}\), confirming our solution.

Key Concepts

Vectors

Vectors are fundamental elements in mathematics and physics that represent quantities with both magnitude and direction. When we talk about vectors in this context, we typically use a notation like \( \mathbf{i} \) and \( \mathbf{j} \), denoting the unit vectors along the x and y axes, respectively. These unit vectors help describe any vector in a plane as a combination of \( \mathbf{i} \) and \( \mathbf{j} \).
For example, the vector \( \mathbf{u} = 2\mathbf{i} + \mathbf{j} \) has a magnitude and a direction.
This vector suggests moving 2 units in the x-direction and 1 unit in the y-direction.

When forming linear combinations, different vectors can be combined using scalar multipliers. This means a vector can be expressed as a sum of others, adjusted by certain scalars.
Understanding this concept helps in analyzing how vectors interact and how they can be composed to form new vectors.

Scalars

Scalars are essential in modifying vectors through multiplication. In a linear combination, scalars are the coefficients that stretch or shrink vectors proportionally. When you have a vector like \( \mathbf{u} \), a scalar \( a \) will alter its magnitude.
Consider the expression \( a \mathbf{v} + b \mathbf{w} \), where \( a \) and \( b \) are scalars and \( \mathbf{v} \) and \( \mathbf{w} \) are vectors.
These scalars determine how much of each vector contributes to the final vector.

• A positive scalar maintains the direction of the vector.
• A negative scalar reverses the direction.
• A zero scalar removes the vector from the combination.

In the solution provided, \( a = \frac{3}{2} \) and \( b = \frac{1}{2} \) ensured the linear combination of the vectors \( \mathbf{v} \) and \( \mathbf{w} \) equal \( \mathbf{u} \.\)

System of Equations

A system of equations is a set of equations with multiple variables that you solve together. In this problem, to find the scalars for a linear combination, you form a system of linear equations from vector definitions.
When you write \( (a + b)\mathbf{i} + (a - b)\mathbf{j} = 2\mathbf{i} + \mathbf{j} \,\), it means that the coefficients of \( \mathbf{i} \) and \( \mathbf{j} \) from the left side of the equation must match those on the right side separately.
This results in the system of equations: \( a + b = 2 \) and \( a - b = 1 \).

To solve, align the equations and employ strategies like addition or substitution.
This simplifies finding values for \( a \) and \( b \) that satisfy both conditions.

• Addition helps eliminate variables by combining equations.
• Substitution replaces one variable with another to solve sequentially.

Here, adding the equations eliminated \( b \,\) allowing us to first find \( a \,\) then using substitution to quickly solve for \( b \.\)