Problem 46 - Contemporary Precalculus

Question

Find a unit vector that has the same direction as \(v\). $$-3 \mathbf{i}-9 \mathbf{j}$$

Step-by-Step Solution

Answer

Answer: The unit vector with the same direction as the vector -3i - 9j is \(-\frac{1}{\sqrt{10}}\mathbf{i} - \frac{3}{\sqrt{10}}\mathbf{j}\).

Step 1: Find the magnitude of the given vector

To find the magnitude of a vector, we can use the formula: \(magnitude = \sqrt {\mathbf{i}^2 + \mathbf{j}^2}\). So for this vector, we have: $$magnitude = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90}$$

Step 2: Divide the given vector by its magnitude

Now that we have the magnitude of the vector, we can divide each component of the vector by that magnitude to find the unit vector. $$\mathbf{\hat{v}} = \frac{\mathbf{v}}{magnitude} = \frac{-3 \mathbf{i} - 9 \mathbf{j}}{\sqrt{90}} = \frac{-3 \mathbf{i}}{\sqrt{90}} + \frac{-9 \mathbf{j}}{\sqrt{90}} = -\frac{1}{\sqrt{10}}\mathbf{i} - \frac{3}{\sqrt{10}}\mathbf{j}$$ So, the unit vector with the same direction as the given vector is \(-\frac{1}{\sqrt{10}}\mathbf{i} - \frac{3}{\sqrt{10}}\mathbf{j}\).

Key Concepts

Vector Magnitude

The magnitude of a vector is a measure of its length and is a crucial concept in understanding vectors. In the context of the exercise, we calculated the magnitude of the vector \( -3 \mathbf{i} - 9 \mathbf{j} \) using the formula \( \sqrt{\mathbf{i}^2 + \mathbf{j}^2} \) where \( \mathbf{i} \) and \( \mathbf{j} \) represent the components of the vector along the x-axis and y-axis, respectively.

To find the magnitude, we squared the components, summed them, and then took the square root of this sum. This gave us \( \sqrt{90} \), which is the length of the vector from the origin to its endpoint in the Cartesian coordinate system. Knowing the magnitude is fundamental before we can talk about normalizing a vector or determining its direction.

Vector Direction

The direction of a vector is the orientation of the arrow that represents the vector, extending from its initial point to its terminal point. In two dimensions, vector direction can be described using the angle it forms with the positive x-axis or by using its x and y components. For the given vector \( -3 \mathbf{i} - 9 \mathbf{j} \), we don't calculate an angle but instead focus on the ratio of its components to represent its direction.

The unit vector derived in the solution, \( -\frac{1}{\sqrt{10}}\mathbf{i} - \frac{3}{\sqrt{10}}\mathbf{j} \), has the same direction as the original vector but has been scaled to a unit length. Despite the vector being rescaled, the ratio of its components (and thus its direction) remains the same. These ratios are key to maintaining the vector's direction in the process of normalization.

Normalizing Vectors

Normalizing a vector involves scaling the vector so that it has a magnitude of 1, without changing its direction. This process creates what is known as a unit vector, symbolized by the hat notation \( \mathbf{\hat{v}} \). The reason for normalizing vectors is often to simplify calculations in physics and engineering, where only the direction of a vector is of interest rather than its magnitude.

As demonstrated in the solution, to normalize the vector \( -3 \mathbf{i} - 9 \mathbf{j} \), we divided each of its components by the vector's magnitude \( \sqrt{90} \). The resulting unit vector \( - \frac{1}{\sqrt{10}}\mathbf{i} - \frac{3}{\sqrt{10}}\mathbf{j} \) represents the direction of the original vector with a standardized length of 1. This ensures that the vector can be easily used for further vectors-related calculations while preserving the vector's original orientation.