Problem 37 - Algebra and Trigonometry

Question

In Exercises \(21-38,\) let $$\mathbf{u}=2 \mathbf{i}-5 \mathbf{j}, \mathbf{v}=-3 \mathbf{i}+7 \mathbf{j}, \text { and } \mathbf{w}=-\mathbf{i}-6 \mathbf{j}$$ Find each specified vector or scalar. $$|\mathbf{w}-\mathbf{u}|$$

Step-by-Step Solution

Answer

The magnitude of the vector w - u is √10.

Step 1: Subtract vectors

First, subtract vector u from vector w to form a new vector. Both vectors are subtracted component-wise. That is: (w1 - u1)i + (w2 - u2)j. Given vectors u = 2i - 5j, and w = -i - 6j; when subtracted we get -1 - 2 = -3 for the i-component and -6 - (-5) = -1 for the j-component. Thus, w - u = -3i - j.

Step 2: Calculate the magnitude

Next, calculate the magnitude (length) of the resultant vector (w-u) using the formula √((w1-u1)² + (w2-u2)²). The magnitude of w - u can be found by evaluating √((-3)² + (-1)²) = √(9+1) = √10.

Key Concepts

Magnitude of a Vector

The magnitude of a vector is an essential concept in algebra and physics.
It measures the length or size of the vector. Think of it as how long the vector is, without considering its direction.

To find the magnitude of a vector, you use the formula:

• If the vector is in the form \(a\mathbf{i} + b\mathbf{j}\), the magnitude is given by \(\sqrt{a^2 + b^2}\).

In our example, after subtracting the vectors, we found that the resulting vector \(\mathbf{w} - \mathbf{u}\) is \(-3\mathbf{i} - \mathbf{j}\).
The magnitude is then \(\sqrt{(-3)^2 + (-1)^2} = \sqrt{10}\).

Understanding magnitude helps you determine how "far" the vector stretches in space, regardless of its direction.

Component-wise Subtraction

Component-wise subtraction is the key to subtracting vectors.
When you subtract one vector from another, you deal with each component separately.

Given vectors \(\mathbf{u} = 2\mathbf{i} - 5\mathbf{j}\) and \(\mathbf{w} = -\mathbf{i} - 6\mathbf{j}\), this means:

• For the \(\mathbf{i}\) components: take \(-1 - 2 = -3\).
• For the \(\mathbf{j}\) components: take \(-6 - (-5) = -1\).

This gives us the new vector \(-3\mathbf{i} - \mathbf{j}\).

Component-wise subtraction makes vector operations straightforward.
It allows you to handle each direction separately, ensuring precision and clarity in calculations.

Vector Operations

Vector operations form a fundamental part of algebra, enabling you to manipulate and analyze vectors effectively.
Operations include addition, subtraction, and finding magnitudes or directions.

In our problem, we focused on subtraction and finding the magnitude.
These are vital tools that help solve physical problems, from calculating forces to determining movement.

By understanding vector operations, you learn how vectors interact, combine, and change in space.
Each operation follows specific rules, just like numbers, but with an added dimension of direction.
This dimensionality is what makes vector operations both challenging and fascinating.

Educational Level: Algebra

Algebra is an exciting branch of mathematics, especially when it comes to vectors.
It's at this educational level that we start solving abstract problems with concrete methods.

Working with vectors in algebra helps build a foundation useful in more advanced fields like calculus and physics.
It provides tools for handling everyday situations where direction and magnitude matter, such as navigation or engineering.

Understanding vectors is not just about solving exercises; it's about opening a window to how the world works.
From calculating distances to analyzing forces, algebra equips you with essential skills for numerous scientific and practical applications.