Problem 44 - Precalculus with Limits

Question

Finding a Unit Vector In Exercises \(39-48,\) find a unit vector in the direction of the given vector. Verify that the result has a magnitude of 1 . $$\mathbf{v}=6 \mathbf{i}-2 \mathbf{j}$$

Step-by-Step Solution

Answer

The unit vector in the direction of the given vector \(\mathbf{v}=6\mathbf{i}-2\mathbf{j}\) is \(\hat{v}=\sqrt{10}/5\mathbf{i}-\sqrt{10}/10\mathbf{j}\)

Step 1: Calculate the Magnitude of the Vector

From the given vector \(\mathbf{v}=6\mathbf{i}-2\mathbf{j}\), we can calculate its magnitude by using the formula: \(\|\mathbf{v}\|=\sqrt{(6^2)+(-2^2)} = 2\sqrt{10}\).

Step 2: Compute the Unit Vector

Now, to find the unit vector, manipulate each component of the original vector by its magnitude. Hence, the unit vector \( \hat{v} = \mathbf{v}/\|\mathbf{v}\|\) gives us \( \hat{v} = (6\mathbf{i}-2\mathbf{j})/(2\sqrt{10}) = \sqrt{10}/5\mathbf{i}-\sqrt{10}/10\mathbf{j}\).

Step 3: Verify the Magnitude

We need to verify that the magnitude of the unit vector is 1. Calculate the magnitude of the unit vector \( \hat{v} = \sqrt{(\sqrt{10}/5)^2 + (-\sqrt{10}/10)^2}\), which simplifies to 1 confirming our solution is correct.

Key Concepts

Vector Magnitude

The magnitude of a vector is a measure of its length or size, and it's the first step in many vector calculations. It is essential for understanding the concept of unit vectors, as the magnitude is used to normalize a vector. To find the magnitude of a vector, we employ the Pythagorean theorem in a coordinate system, summing the squares of all components and taking the square root of this sum.

For a two-dimensional vector like \( \mathbf{v}=6\mathbf{i}-2\mathbf{j} \), the magnitude is calculated using the formula \( \|\mathbf{v}\|=\sqrt{(6^2)+(-2^2)} \) which yields \( 2\sqrt{10} \). In this context, \( \mathbf{i} \) and \( \mathbf{j} \) are the unit vectors along the x and y axes, respectively, scaled by the corresponding component of the vector \( \mathbf{v} \). The magnitude tells us the 'length' of the vector from the origin point to its end point in the vector space.

Vector Direction

A vector is not just a numeric value; it also has a direction, which is a crucial part of its definition. The direction is determined by the vector's components and describes where the vector is pointing in space. For instance, the vector \( \mathbf{v}=6\mathbf{i}-2\mathbf{j} \) implies a direction primarily along the x-axis, due to the larger x-component (6), and slightly in the negative y-direction because of the negative y-component (-2).

Direction can be described by the angle a vector makes with a reference axis, but in many mathematical treatments, the exact angular direction isn't necessary. Instead, what is often required is the directional proportions of the vector—the ratio of each component to the magnitude. When finding a unit vector, the direction of the original vector is preserved, while its magnitude is scaled to 1, thereby maintaining the vector's orientation in space.

Unit Vector Calculation

Creating a unit vector involves scaling a vector so that it has a magnitude of 1, but its direction remains unchanged. A unit vector is often denoted by a hat, as in \( \hat{v} \). To calculate a unit vector, you divide the original vector by its magnitude.

For \( \mathbf{v}=6\mathbf{i}-2\mathbf{j} \), the unit vector is computed as follows: \( \hat{v} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \) which gives us \( \hat{v} = \frac{6\mathbf{i}-2\mathbf{j}}{2\sqrt{10}} \). Simplify the components to get \( \hat{v} = \frac{\sqrt{10}}{5}\mathbf{i}-\frac{\sqrt{10}}{10}\mathbf{j} \). It's vital to check your result, verifying the magnitude of the unit vector is indeed 1, confirming that while its length has been normalized, its direction proportional to the original vector has been preserved.