Problem 18 - Precalculus

Question

Assume that the vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c},\) and \(\mathbf{d}\) are defined as follows: $$\mathbf{a}=\langle 2,3\rangle \quad \mathbf{b}=\langle 5,4\rangle \quad \mathbf{c}=\langle 6,-1\rangle \quad \mathbf{d}=\langle-2,0\rangle$$ Compute each of the indicated quantities. $$|5 \mathbf{b}+5 \mathbf{c}|$$

Step-by-Step Solution

Answer

|5\mathbf{b}+5\mathbf{c}| \approx 57.0

Step 1: Multiply Vectors by Scalar

First, we need to find the vectors \(5 \mathbf{b}\) and \(5 \mathbf{c}\). This means multiplying each component of \(\mathbf{b}\) and \(\mathbf{c}\) by 5. For \(5 \mathbf{b}\): \(5 \times 5 = 25\) and \(5 \times 4 = 20\). So, \(5 \mathbf{b} = \langle 25, 20 \rangle\). For \(5 \mathbf{c}\): \(5 \times 6 = 30\) and \(5 \times -1 = -5\). So, \(5 \mathbf{c} = \langle 30, -5 \rangle\).

Step 2: Add the Vectors

Next, we add the vectors \(5 \mathbf{b}\) and \(5 \mathbf{c}\). Add the corresponding components: \(25 + 30 = 55\) and \(20 + (-5) = 15\). Thus, \(5 \mathbf{b} + 5 \mathbf{c} = \langle 55, 15 \rangle\).

Step 3: Calculate Magnitude

The magnitude of a vector \(\mathbf{v} = \langle x, y \rangle\) is calculated using the formula: \(|\mathbf{v}| = \sqrt{x^2 + y^2}\). For the vector \(\langle 55, 15 \rangle\): First, square the components: \(55^2 = 3025\) and \(15^2 = 225\). Then, add them: \(3025 + 225 = 3250\). Finally, take the square root: \(|5 \mathbf{b} + 5 \mathbf{c}| = \sqrt{3250}\).

Step 4: Simplify the Magnitude

Calculate the numerical value of \(\sqrt{3250}\) for the final answer. This equals approximately \( 57.0 \).

Key Concepts

Vector Scalar Multiplication

In vector algebra, one of the key operations is scalar multiplication. This operation involves multiplying a vector by a scalar (a constant). For any vector, each of its components is multiplied by the scalar to form a new vector.

For example, if we have a vector \( \mathbf{b} = \langle x, y \rangle \), and a scalar \( k \), then the scalar multiplication \( k \mathbf{b} \) results in a new vector \( \langle kx, ky \rangle \). In the exercise, scalar multiplication was applied to the vectors \( \mathbf{b} \) and \( \mathbf{c} \) as follows:

• \( 5 \times \mathbf{b} = \langle 25, 20 \rangle \)
• \( 5 \times \mathbf{c} = \langle 30, -5 \rangle \)

This shows how each component of the vector is scaled up by a factor of 5. Scalar multiplication is essential in scaling vectors and is often a preliminary step in more complex calculations involving vectors.

Vector Addition

Vector addition is another fundamental concept in vector algebra. To add two vectors, you simply add their corresponding components. If you have the vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), their sum \( \mathbf{u} + \mathbf{v} \) is the new vector \( \langle u_1 + v_1, u_2 + v_2 \rangle \).

In the exercise, after obtaining the scaled vectors \( 5\mathbf{b} \) and \( 5\mathbf{c} \), we added them to get another vector:

• For the first components: \( 25 + 30 = 55 \)
• For the second components: \( 20 + (-5) = 15 \)

This resulted in the vector \( \langle 55, 15 \rangle \). Vector addition is paramount when combining forces, directions, or any quantities that have both magnitude and direction.

Vector Magnitude

The magnitude of a vector measures its length or size, a crucial concept in understanding vector properties. The magnitude of a vector \( \mathbf{v} = \langle x, y \rangle \) is calculated using the Pythagorean theorem as \(|\mathbf{v}| = \sqrt{x^2 + y^2}\). It provides a way to quantify how large or small the vector is.

For the resulting vector from our exercise, \( \langle 55, 15 \rangle \), the magnitude is determined by:

• Calculating \( 55^2 \), which equals 3025
• Calculating \( 15^2 \), which equals 225
• Summing these values: 3025 + 225 = 3250
• Taking the square root: \( \sqrt{3250} \approx 57.0 \)

The magnitude provides insight into the overall scale of the vector, broadening our understanding of its physical significance.

Precalculus Problems

Precalculus problems involving vectors often combine concepts like vector scalar multiplication, addition, and magnitude calculation. These problems help set the foundation for higher-level math courses, making them essential building blocks.

In tackling precalculus problems, students learn not just how to compute, but also to see the interplay between mathematical operations and their geometric interpretations. The exercise demonstrated a sequence of solving complex math problems step-by-step, aiding students in developing problem-solving skills crucial for calculus and beyond.

• Identifying components for scalar multiplication
• Executing vector addition to find resultant vectors
• Calculating and simplifying magnitudes

Solving these problems enhances understanding and builds confidence when approaching rigorous and abstract math in future studies.