Problem 46 - Algebra and Trigonometry

Question

In Exercises \(39-46\), find the unit vector that has the same direction as the vector \(\mathbf{v}\) $$\mathbf{v}=\mathbf{i}-\mathbf{j}$$

Step-by-Step Solution

Answer

The unit vector that has the same direction as the vector \(\mathbf{v}=\mathbf{i}-\mathbf{j}\) is \(\mathbf{u}=\frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}\)

Step 1: Identify the components of the vector

The vector is given as \(\mathbf{v}=\mathbf{i}-\mathbf{j}\), which means the components are \(x=1\) and \(y=-1\).

Step 2: Calculate the magnitude of the vector

The magnitude or length of a vector can be calculated using the Pythagorean Theorem. The formula for the magnitude of a vector in two dimensions is \(\|v\|=\sqrt{x^2 + y^2}\). So for this vector, the magnitude is \(\sqrt{1^2 + (-1)^2}=\sqrt{2}\).

Step 3: Divide each component by magnitude

To get the unit vector, each component of the original vector should be divided by the magnitude. This gives the unit vector as \(\mathbf{u}=\frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}\)

Key Concepts

Vector Magnitude

When dealing with vectors in mathematics and physics, understanding the concept of vector magnitude is crucial. Simply put, the magnitude of a vector refers to its 'length' in the spatial context it occupies, whether that be two-dimensional or three-dimensional space. To find the magnitude, we turn to the geometrical foundation laid by the Pythagorean theorem.

For instance, if we have a vector \textbf{v} with components (x, y) in a two-dimensional space, its magnitude, denoted as \(\|v\|\), is given by \(\|v\| = \sqrt{x^2 + y^2}\). Think of it as measuring the hypotenuse of a right-angled triangle, with x and y being the perpendicular sides. The magnitude is always a positive number or zero, and it essentially reflects the vector's 'size' irrespective of its direction.

Applying this principle to the exercise, we had a vector \textbf{v} with components (1, -1), leading to a magnitude of \(\sqrt{1^2 + (-1)^2} = \sqrt{2}\), which is a crucial step in gaining insight into a vector's properties, especially when further manipulation, like normalization to a unit vector, is required.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in mathematics that establishes a relationship between the lengths of the sides of a right triangle. For any right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

This theorem can be formally stated as \(c^2 = a^2 + b^2\), where \(c\) is the length of the hypotenuse while \(a\) and \(b\) are the lengths of the other two sides. In vector calculus, this theorem plays a vital role in calculating vector magnitudes. The components of a vector in Cartesian space can be treated as sides of a right triangle, with the vector itself representing the hypotenuse. Thus, determining the magnitude of any two-dimensional vector resorts back to this classic theorem.

In the case of our exercise, by applying the theorem (\(\|v\| = \sqrt{x^2 + y^2}\)), we take our components \(x=1\) and \(y=-1\), and find the magnitude, paving the way for the unit vector calculation which will ultimately express our vector in its simplest directional form.

Vector Components

Vector components are the influences that a vector has along the axes of the coordinate system in which it is defined. Typically, in a two-dimensional Cartesian system, these components are along the X-axis and Y-axis, denoted as \(x\) and \(y\) respectively. They can be viewed as the projections of the vector along these axes, telling us how far along each axis the vector reaches.

In a practical setting, if you're given a vector \(\mathbf{v}\), it can be expressed in terms of its components as \(\mathbf{v} = x\mathbf{i} + y\mathbf{j}\), with \(\mathbf{i}\) and \(\mathbf{j}\) being the unit vectors along the X-axis and Y-axis respectively. Essentially, vector components allow us to break down a vector into simpler parts that can be easily worked with, especially when calculating vector magnitudes or when adding vectors together.

Regarding our exercise’s vector \(\mathbf{v} = \mathbf{i} - \mathbf{j}\), its components were determined as \(x = 1\) and \(y = -1\), serving as a starting point for calculating the vector's magnitude, and ultimately for finding the unit vector that retains the original vector's direction.