Problem 53 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find a unit vector in the direction of the given vector. Verify that the result has a magnitude of 1.$$\mathbf{v}=4 \mathbf{i}-3 \mathbf{j}$$.

Step-by-Step Solution

Answer

The unit vector in the direction of the given vector \( \mathbf{v}= 4\mathbf{i} - 3\mathbf{j} \) is \( \mathbf{u} = (4/5)\mathbf{i} - (3/5)\mathbf{j}\), and its magnitude is 1.

Step 1: Find the Magnitude of the Given Vector

Firstly, find the magnitude or the length of the given vector \(\mathbf{v}= 4\mathbf{i} - 3\mathbf{j}\). The magnitude of a vector is given by the square root of the sum of the squares of its components. In this case, it's \( \|\mathbf{v}\| = \sqrt{(4^2 + (-3)^2)} = \sqrt{16 + 9} = \sqrt{25} = 5 \).

Step 2: Calculate the Unit Vector

Secondly, find the unit vector in the direction of the given vector. This is done by dividing each component of the given vector by its magnitude. Therefore, the unit vector \( \mathbf{u} \) is given by \( \mathbf{u} = \mathbf{v}/ \|\mathbf{v}\| = (4/5)\mathbf{i} - (3/5)\mathbf{j} \).

Step 3: Verify the Magnitude of the Unit Vector

Lastly, verify that the length or the magnitude of the unit vector calculated is indeed 1. Find the magnitude of the unit vector \( \mathbf{u} \) by squaring the components, adding them and taking the square root. So, \( \|\mathbf{u}\| = \sqrt{((4/5)^2 + (-3/5)^2)} = \sqrt{0.64 + 0.36} = \sqrt{1} = 1 \). This confirms that the magnitude of \(\mathbf{u} \) is indeed 1.

Key Concepts

Vector Magnitude

The concept of vector magnitude is essential for understanding vectors in mathematics and physics. Think of it as the length of the vector. For a 2D vector, like the one given in the exercise \(\mathbf{v} = 4\mathbf{i} - 3\mathbf{j}\), the magnitude represents the distance from the origin to the point \( (4, -3) \) in the Cartesian coordinate system.

To find the vector's magnitude, you use the formula:

• Square each component of the vector.
• Add these squares together.
• Take the square root of the sum.

This is expressed as \( \|\mathbf{v}\| = \sqrt{(a^2 + b^2)} \) for a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \).
The calculation for the given vector yields \( \sqrt{16 + 9} = 5 \). This means the vector has a magnitude of 5.

Knowing the magnitude allows you to understand the size of the vector without considering its direction.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector. A unit vector has a magnitude of 1, but it retains the direction of the original vector. It essentially "scales down" the vector, keeping its essential direction but changing its length to 1.

To normalize a vector, use the following steps:

• Calculate the magnitude of the vector as discussed.
• Divide each component of the vector by this magnitude.

For the vector \(\mathbf{v} = 4\mathbf{i} - 3\mathbf{j}\), the magnitude is 5. Thus, the unit vector \(\mathbf{u}\) is found by dividing each component by 5:
\(\mathbf{u} = \frac{4}{5}\mathbf{i} - \frac{3}{5}\mathbf{j}\).

This gives us a vector in the same direction as \(\mathbf{v}\), but now with a magnitude of 1, confirming it is indeed a unit vector.

Vector Components

The components of a vector are the projections of that vector along the axes of the coordinate system. In a 2D space, for a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), \( a \) and \( b \) are its components along the x and y axes, respectively.

Components are important because they help define the direction of a vector. They show how much of the vector is pointing in each direction of the x and y axis. For our vector \( \mathbf{v} = 4\mathbf{i} - 3\mathbf{j} \), the components are 4 in the x-direction and -3 in the y-direction.

Understanding vector components is crucial for operations like addition, subtraction, and especially for deriving the unit vector, as you are distributing these components by the magnitude to obtain a normalized direction. This process reveals the underlying force or velocity as distributed across dimensions.