Problem 43 - University Calculus: Early Transcendentals

Question

Linear combination \(\quad\) Let \(\mathbf{u}=\langle 1,2,1\rangle, \mathbf{v}=\langle 1,-1,-1\rangle\) \(\mathbf{w}=\langle 1,1,-1\rangle,\) and \(\mathbf{z}=\langle 2,-3,-4\rangle .\) Find scalars \(a, b,\) and \(c\) such that \(\mathbf{z}=a \mathbf{u}+b \mathbf{v}+c \mathbf{w}\).

Step-by-Step Solution

Answer

Scalars are \(a = -1\), \(b = 2\), \(c = 1\).

Step 1: Identify the Linear Combination

We know that to find the scalars \(a, b,\) and \(c\), we need to solve the equation \(\mathbf{z} = a \mathbf{u} + b \mathbf{v} + c \mathbf{w}\). Written in component form: \[ \langle 2, -3, -4 \rangle = a \langle 1, 2, 1 \rangle + b \langle 1, -1, -1 \rangle + c \langle 1, 1, -1 \rangle \].

Step 2: Set Up the System of Equations

Equate corresponding components from both sides of \(\mathbf{z} = a \mathbf{u} + b \mathbf{v} + c \mathbf{w}\):\[1. \quad a + b + c = 2\] \[2. \quad 2a - b + c = -3\] \[3. \quad a - b - c = -4\]

Step 3: Use Equation 1 to Express c

From Equation 1, we have: \(c = 2 - a - b\). Substitute this expression for \(c\) into Equations 2 and 3.

Step 4: Substitute c into Equation 2

Replace \(c\) in Equation 2: \(2a - b + (2 - a - b) = -3\). Simplify to get: \(a - 2b = -5\). Let's call this Equation 4.

Step 5: Substitute c into Equation 3

Replace \(c\) in Equation 3: \(a - b - (2 - a - b) = -4\). Simplify to get: \(2a = -2\). So, \(a = -1\).

Step 6: Solve for b Using Equation 4

Substitute \(a = -1\) into Equation 4: \(-1 - 2b = -5\). Simplify to find \(b = 2\).

Step 7: Solve for c

Now that we know \(a = -1\) and \(b = 2\), substitute these into the expression for \(c\): \(c = 2 - (-1) - 2 = 1\).

Step 8: Verify the Solution

To verify, substitute \(a = -1\), \(b = 2\), \(c = 1\) back into the original vector equation and ensure equality: \((-1)\langle 1,2,1 \rangle + 2\langle 1,-1,-1 \rangle + 1\langle 1,1,-1 \rangle = \langle 2,-3,-4 \rangle\). After simplifying, it holds true.

Key Concepts

Vectors

Vectors are mathematical objects that have both a magnitude and a direction. They are often represented in the coordinate form as an ordered list of numbers, such as \( \langle 1, 2, 1 \rangle \). Each number in this list is called a component of the vector. Vectors can be visualized as arrows in space, where the length of the arrow represents the magnitude and the direction in which the arrow points represents the direction of the vector.
Understanding vectors is essential in fields like physics, engineering, and computer science, as they represent quantities such as force, velocity, and displacement.

• Magnitude: The length of the vector, calculated as \( \sqrt{x_1^2 + x_2^2 + x_3^2} \) for a 3D vector.
• Direction: The angle or orientation of the vector in space.

Operations on vectors, such as addition, subtraction and scalar multiplication, form the basis of vector calculus and are key in solving linear combinations.

System of Equations

A system of equations is a set of two or more equations that share a common set of unknowns. In the context of vectors and linear combinations, it involves solving for unknown scalars that, when applied to given vectors, can produce a specified vector.
To solve a system of equations, we need as many equations as there are unknowns. For instance, in our exercise, we set up three equations to solve for the three unknowns: \(a\), \(b\), and \(c\). These equations come from equating the components of the linear combination of vectors to the components of the vector \(\mathbf{z}\) that we are responding to.

• Substitution: Replace one variable with an expression from another equation.
• Elimination: Add or subtract equations to eliminate a variable.

Solving systems of equations is a fundamental skill in algebra that helps solve many real-world problems, such as determining the points that satisfy multiple conditions.

Vector Operations

Vector operations are mathematical processes that can be performed on vectors to change them or combine them in useful ways. The operations usually discussed include addition, subtraction, and multiplication by a scalar.
Addition and subtraction of vectors are performed component-wise, meaning each component of one vector is added to or subtracted from the respective component of another vector. This results in a new vector. Scalar multiplication involves multiplying each component of a vector by a scalar value, effectively changing its magnitude without altering its direction.

• Vector Addition: \(\langle 1, 2, 1 \rangle + \langle 1, -1, -1 \rangle = \langle 2, 1, 0 \rangle \).
• Scalar Multiplication: \( 3 \times \langle 1, 2, 1 \rangle = \langle 3, 6, 3 \rangle \).

Understanding and applying these operations allows for the manipulation of vectors in various dimensions, which is crucial in fields such as physics to compute net forces and in computer graphics for transformations.