Problem 9 - 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 vectors are orthogonal because their dot product is 0.

Step 1: Find the dot product

The dot product of two vectors \( \mathbf{v} \) and \( \mathbf{w} \) is calculated using the formula \[ \mathbf{v} \cdot \mathbf{w} = v_1w_1 + v_2w_2 \] Here, we have \( \mathbf{v} = \mathbf{i} - \mathbf{j} \) and \( \mathbf{w} = \mathbf{i} + \mathbf{j} \). \ v_1 = 1, v_2 = -1, w_1 = 1, w_2 = 1. Therefore, \ \[ \mathbf{v} \cdot \mathbf{w} = (1)(1) + (-1)(1) = 1 - 1 = 0 \]

Step 2: Find the magnitudes of the vectors

The magnitude of a vector \( \mathbf{v} \) is given by \[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2} \] \ \ \[ |\mathbf{v}| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \] Similarly, for \( \mathbf{w} \): \[ |\mathbf{w}| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

Step 3: Find the angle between the vectors

The angle \( \theta \) between two vectors can be found using the formula \[ \cos \( \theta \) = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}| |\mathbf{w}|} \] Substituting the values from Steps 1 and 2: \[ \cos \( \theta \) = \frac{0}{ \sqrt{2} \cdot \sqrt{2}} = 0 \] Therefore, \( \theta = \arccos(0) = \frac{\pi}{2} \) radians or 90 degrees.

Step 4: Determine the relationship between the vectors

Since the dot product is 0, the vectors \( \mathbf{v} \) and \( \mathbf{w} \) are orthogonal.

Key Concepts

dot product

In vector algebra, the dot product (also known as the scalar product) is an operation that takes two vectors and returns a single real number. The dot product of two vectors \(\mathbf{v} \) and \(\mathbf{w} \) is calculated using the formula:

\[ \mathbf{v} \cdot \mathbf{w} = v_1w_1 + v_2w_2 \]

Here, \( \mathbf{v} \) and \( \mathbf{w} \) are expressed in terms of their components. So, if \( \mathbf{v} = \mathbf{i} - \mathbf{j} \) and \( \mathbf{w} = \mathbf{i} + \mathbf{j} \), this means \( v_1 = 1 \), \( v_2 = -1 \), \( w_1 = 1 \), and \( w_2 = 1 \).

Substituting these values, we get:

\[ \mathbf{v} \cdot \mathbf{w} = (1)(1) + (-1)(1) = 1 - 1 = 0 \]

The result is 0, indicating an important relationship between the vectors which we will explore further in upcoming sections.

vector magnitude

The magnitude of a vector, also known as its length or norm, indicates how long the vector is. For a vector \( \mathbf{v} = (v_1, v_2) \), the magnitude is calculated as:

\[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2} \]

Let's compute the magnitudes of our vectors. For \( \mathbf{v} = \mathbf{i} - \mathbf{j} \)

\[ |\mathbf{v}| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

Similarly, for \( \mathbf{w} = \mathbf{i} + \mathbf{j} \)

\[ |\mathbf{w}| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

Both vectors have the same magnitude of \( \sqrt{2} \). Knowing the magnitude is essential for further calculations such as finding the angle between vectors.

angle between vectors

The angle between two vectors can help determine the geometric relationship between them. The cosine of the angle \(\theta\) between vectors \(\mathbf{v} \) and \(\mathbf{w} \) is given by:

\[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}| |\mathbf{w}|} \]

We previously found that \( \mathbf{v} \cdot \mathbf{w} = 0 \) and both \( |\mathbf{v}| \) and \( |\mathbf{w}| \) are \( \sqrt{2} \). Substituting these into our formula, we get:

\[ \cos(\theta) = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = 0 \]

Since \( \cos(\theta) = 0 \), \(\theta\) must be \( \frac{\pi}{2} \) radians or 90 degrees. This shows that the vectors are perpendicular to each other, meaning they intersect at a right angle.

orthogonal vectors

Two vectors are orthogonal (or perpendicular) if their dot product is zero. This means that the angle between them is 90 degrees.

Considering our vectors \( \mathbf{v} \) and \( \mathbf{w} \):

* We found \( \mathbf{v} \cdot \mathbf{w} = 0 \)

* We calculated the angle between them to be 90 degrees

With these results, we can confidently state that the vectors \( \mathbf{v} \) and \( \mathbf{w} \) are orthogonal. Orthogonality is a critical concept in vector algebra, as it often indicates independence between vectors, which is useful in numerous applications, including physics and engineering.