Question

Let \(\mathbf{u}=\langle 3,-4\rangle, \mathbf{v}=\langle 1,1\rangle,\) and \(\mathbf{w}=\langle-1,0\rangle .\) Carry out the following computations. Which has the greater magnitude, \(2 \mathbf{u}\) or \(7 \mathbf{v} ?\)

Short answer

Answer: In this problem, the magnitude of \(2\mathbf{u}\) is greater than the magnitude of \(7\mathbf{v}\) (10 > 9.9).

Step-by-step solution

Compute the scalar multiplications of \(\mathbf{u}\) and \(\mathbf{v}\)

Recall that to perform scalar multiplication, we multiply each component of the vector by the scalar. We have: \(2\mathbf{u} = 2\langle 3, -4\rangle = \langle (2)(3), (2)(-4)\rangle = \langle 6, -8\rangle\) \(7\mathbf{v} = 7\langle 1, 1\rangle = \langle (7)(1), (7)(1)\rangle = \langle 7, 7\rangle\) Now we have obtained \(2\mathbf{u} = \langle 6, -8\rangle\) and \(7\mathbf{v} = \langle 7, 7\rangle\).

Calculate the magnitudes of \(2\mathbf{u}\) and \(7\mathbf{v}\)

The magnitude of a vector \(\langle a,b\rangle\) is given by the formula: \(||\langle a, b \rangle|| = \sqrt{a^2 + b^2}\) Using this formula, compute the magnitudes of the vectors \(2\mathbf{u}\) and \(7\mathbf{v}\): Magnitude of \(2\mathbf{u}\): \(||\langle 6, -8 \rangle|| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\) Magnitude of \(7\mathbf{v}\): \(||\langle 7, 7 \rangle|| = \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} \approx 9.9\)

Compare magnitudes

After comparing the magnitudes of both scalar multiplications, we have: \(||2\mathbf{u}|| = 10\) \(||7\mathbf{v}|| \approx 9.9\) Since the magnitude of \(2\mathbf{u}\) is greater than the magnitude of \(7\mathbf{v}\), we can conclude that \(2\mathbf{u}\) has the greater magnitude.

Key concepts

Scalar Multiplication

Scalar multiplication is an essential operation in vector mathematics. It involves multiplying a vector by a scalar (a constant number). This operation scales the vector by the given scalar value, affecting each component of the vector equally.

To perform scalar multiplication, simply multiply every component of the vector by the scalar. In this exercise:

• We take vector \(\mathbf{u} = \langle 3, -4 \rangle\) and multiply it by scalar 2. This gives us \(2\mathbf{u} = \langle 6, -8 \rangle\).
• Similarly, vector \(\mathbf{v} = \langle 1, 1 \rangle\) when multiplied by scalar 7 becomes \(7\mathbf{v} = \langle 7, 7 \rangle\).

This step transforms the vectors into larger or smaller versions that maintain the direction of the original vector. Thus, scalar multiplication plays a key role in resizing vectors while preserving direction.

Magnitude Comparison

Comparing magnitudes of vectors is used to determine which of them is longer or larger. The magnitude of a vector in two dimensions \( \langle a, b \rangle \) is given by the formula \( ||\langle a, b \rangle|| = \sqrt{a^2 + b^2} \). This magnitude represents the length of the vector from the origin to the point \( \langle a, b \rangle \) in space.

For our exercise, we calculated:

• The magnitude of \(2\mathbf{u} = \langle 6, -8 \rangle\) as \( \sqrt{6^2 + (-8)^2} = 10 \).
• The magnitude of \(7\mathbf{v} = \langle 7, 7 \rangle\) as \( \sqrt{7^2 + 7^2} \approx 9.9 \).

By comparing these results, it's evident that the scalar-multiplied vector \(2\mathbf{u}\) has a slightly greater length than \(7\mathbf{v}\). This highlights how scalar multiplication can affect the magnitude of vectors, and why understanding this concept is important when analyzing vector relationships.

Vector Operations

Vector operations involve actions such as addition, subtraction, and scalar multiplication. These operations can be performed between vectors or between vectors and scalars to manipulate vector properties like direction and magnitude.

Key operations include:

• Addition: Adding vectors involves adding their corresponding components, resulting in a new vector. For example, adding vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\) results in \(\langle a_1 + b_1, a_2 + b_2 \rangle\).
• Subtraction: Similar to addition, but instead you subtract the components. So, \(\mathbf{a} - \mathbf{b} = \langle a_1 - b_1, a_2 - b_2 \rangle\).
• Scalar Multiplication: We've already discussed scalar multiplication, which changes the size of the vector based on the scalar value.

These operations allow us to model and solve problems in physics and engineering by providing a way to easily manipulate quantities that have both magnitude and direction. Understanding vector operations is crucial for solving real-world problems efficiently.