Problem 10 - Algebra and Trigonometry

Question

(a) find the dot product v \(\cdot \mathbf{w} ;\) (b) find the angle between \(\mathbf{v}\) and \(\mathbf{w} ;\) (c) state whether the vectors are parallel, orthogonal, or neither. $$ \mathbf{v}=\mathbf{i}+\mathbf{j}, \quad \mathbf{w}=-\mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

The dot product is 0; the angle is \(90^\circ\); the vectors are orthogonal.

Step 1 - Understand the Vectors

We are given two vectors: \(\mathbf{v}=\mathbf{i}+\mathbf{j}\text{ and } \mathbf{w}=-\mathbf{i}+\mathbf{j}\). The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) correspond to the x and y components respectively.

Step 2 - Find the Dot Product

The dot product of two vectors \(\mathbf{v}\) and \(\mathbf{w}\) is computed as follows: \( \mathbf{v} \cdot \mathbf{w} = (1)(-1) + (1)(1)\). Thus, \( \mathbf{v} \cdot \mathbf{w} = -1 + 1 = 0\).

Step 3 - Calculate the Magnitudes

Find the magnitudes of the vectors \(\mathbf{v}\) and \(\mathbf{w}\): \(\|\mathbf{v}\| = \sqrt{(1)^2 + (1)^2} = \sqrt{2}\) and \(\|\mathbf{w}\| = \sqrt{(-1)^2 + (1)^2} = \sqrt{2}\).

Step 4 - Determine the Angle Between the Vectors

Use the dot product and magnitudes to find the cosine of the angle \(\theta\) between the vectors. The formula is: \( \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|}\). Since \( \mathbf{v} \cdot \mathbf{w} = 0\) and \( \|\mathbf{v}\| \|\mathbf{w}\| = \sqrt{2} \times \sqrt{2} = 2\), we have \( \cos(\theta) = \frac{0}{2} = 0\). Thus, \( \theta = \cos^{-1}(0) = \frac{\pi}{2}\) radians (or \(90^\circ\)).

Step 5 - Classify the Vectors

Two vectors are orthogonal if their dot product is zero. Since \( \mathbf{v} \cdot \mathbf{w} = 0\), \(\mathbf{v}\) and \(\mathbf{w}\) are orthogonal.

Key Concepts

vector algebra

In vector algebra, vectors are mathematical objects characterized by both direction and magnitude. They are commonly represented in Cartesian coordinates as combinations of unit vectors, such as \(\textbf{i}\) for the x-axis and \(\textbf{j}\) for the y-axis. For instance, in the given exercise, we have two vectors \(\textbf{v} = \textbf{i} + \textbf{j}\) and \(\textbf{w} = -\textbf{i} + \textbf{j}\). Here, \(\textbf{i}\) and \(\textbf{j}\) indicate the respective components along the x-axis and y-axis.

Vectors can be added, subtracted, and scaled by multiplying by a scalar. In addition, the dot product is a key operation in vector algebra that combines two vectors to produce a scalar. This scalar can provide crucial information regarding the relationship between the vectors, such as their orthogonality or the angle between them.

magnitude of vectors

The magnitude of a vector, denoted as \(\big|\textbf{v}\big|\), measures its length or size. To compute the magnitude of a vector, we rely on the Pythagorean theorem. For a vector \(\textbf{v} = a\textbf{i} + b\textbf{j}\), the magnitude is calculated using the formula:

\(\big|\textbf{v}\big| = \sqrt{a^2 + b^2}\)

Applying this to our vectors in the exercise:

• The magnitude of \(\textbf{v} = \textbf{i} + \textbf{j}\) is \(\big|\textbf{v}\big| = \sqrt{1^2 + 1^2} = \sqrt{2}\).
• Similarly, the magnitude of \(\textbf{w} = -\textbf{i} + \textbf{j}\) is \(\big|\textbf{w}\big| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}\).

Understanding the magnitudes is essential for calculating the angle between vectors and assessing other vector properties.

orthogonal vectors

Vectors are said to be orthogonal if they are perpendicular to each other. One of the most straightforward methods to determine whether two vectors are orthogonal is to compute their dot product. If the dot product of two vectors equals zero, then the vectors are orthogonal. In our exercise:

\(\textbf{v} \cdot \textbf{w} = (1)(-1) + (1)(1) = -1 + 1 = 0\)

Because the dot product is zero, we can conclude that the vectors \(\textbf{v}\) and \(\textbf{w}\) are orthogonal. Orthogonal vectors play a vital role in various fields, such as physics and engineering, where right angles are crucial for many applications.

angle between vectors

The angle between two vectors, \(\theta\), can be found using the dot product formula:

\(\text{cos}(\theta) = \frac{\textbf{v} \cdot \textbf{w}}{\big|\textbf{v}\big| \big|\textbf{w}\big|}\)

From our exercise, we know:

• \(\textbf{v} \cdot \textbf{w} = 0\)
• \(\big|\textbf{v}\big| = \sqrt{2}\)
• \(\big|\textbf{w}\big| = \sqrt{2}\)

Substituting these values into the formula gives:

\(\text{cos}(\theta) = \frac{0}{\big|\textbf{v}\big| \big|\textbf{w}\big|} = \frac{0}{\big|\textbf{v}\big| \times \big|\textbf{w}\big|} = \frac{0}{2} = 0\)

Therefore, \(\theta = \text{cos}^{-1}(0) = \frac{\text{\textpi}}{2}\) radians or 90 degrees. This confirms that \(\textbf{v}\) and \(\textbf{w}\) are orthogonal, as the angle between them is 90 degrees.