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. $$ \mathbf{v}=2 \mathbf{i}+2 \mathbf{j}, \quad \mathbf{w}=\mathbf{i}+2 \mathbf{j} $$

Short answer

Dot product: 6; Angle: \( \theta = \cos^{-1}\left(\frac{3 \sqrt{10}}{10}\right) \); Vectors are neither parallel nor orthogonal.

Step-by-step solution

Write the given vectors

The given vectors are: \( \mathbf{v} = 2\mathbf{i} + 2\mathbf{j} \) and \( \mathbf{w} = \mathbf{i} + 2\mathbf{j} \).

Find the dot product

The dot product \( \mathbf{v} \cdot \mathbf{w} \) is calculated as follows:\[ \mathbf{v} \cdot \mathbf{w} = (2)(1) + (2)(2) = 2 + 4 = 6 \].

Calculate magnitudes of the vectors

The magnitudes of the vectors are: \( \|\mathbf{v}\| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2} \)and \( \|\mathbf{w}\| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \).

Find the cosine of the angle

Use the dot product formula \( \mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos(\theta) \): \[ 6 = (2 \sqrt{2})(\sqrt{5}) \cos(\theta) \] Solving for \( \cos(\theta) \): \[ \cos(\theta) = \frac{6}{2 \sqrt{2} \cdot \sqrt{5}} = \frac{6}{2 \sqrt{10}} = \frac{3}{\sqrt{10}} \] \[ \cos(\theta) = \frac{3\sqrt{10}}{10} \].

Determine the angle \( \theta \)

The angle \( \theta \) is given by: \[ \theta = \cos^{-1}\left(\frac{3 \sqrt{10}}{10}\right) \].

Conclude whether vectors are parallel, orthogonal or neither

Vectors are parallel if \( \cos(\theta) = 1 \) or \( \cos(\theta) = -1 \) (\( \theta = 0 \)° or 180°), and orthogonal if \( \cos(\theta) = 0 \) (\( \theta = 90 \)°).Here, \( \cos(\theta) = \frac{3 \sqrt{10}}{10} \), which is neither 1, -1, nor 0, so the vectors are neither parallel nor orthogonal.

Key concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation between two vectors. Unlike the cross product, which results in a vector, the dot product yields a scalar. To compute the dot product of two vectors, you multiply corresponding components and then sum these products. Let's break down the process with an example where {\( \mathbf{v} = 2\mathbf{i} + 2\mathbf{j} \)} and {\( \mathbf{w} = \mathbf{i} + 2\mathbf{j} \)}.

The formula for the dot product is:

$$ \mathbf{v} \cdot \mathbf{w} = v_1w_1 + v_2w_2 $$

Here, our components are:

• {\( v_1 = 2, v_2 = 2 \)}
• {\( w_1 = 1, w_2 = 2 \)}

Using the formula, we find:

$$ \mathbf{v} \cdot \mathbf{w} = (2)(1) + (2)(2) = 2 + 4 = 6 $$

Thus, the dot product of \({\mathbf{v}}\) and \({\mathbf{w}}\) is 6.

Vector Magnitudes

The magnitude of a vector, often called its length or norm, is a measure of how long the vector is. To find the magnitude of a vector, you can use the Pythagorean theorem. The magnitude of a vector {\( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} \)} is given by:

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

For the vectors {\( \mathbf{v} = 2\mathbf{i} + 2\mathbf{j} \)} and {\( \mathbf{w} = \mathbf{i} + 2\mathbf{j} \)}, their magnitudes are calculated as follows:

• Magnitude of \({\mathbf{v}}\):
  \|\mathbf{v}\| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
• Magnitude of \({\mathbf{w}}\): \|\mathbf{w}\| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}

Understanding vector magnitudes is essential for calculating the angle between vectors.

Angle Between Vectors

The angle between two vectors reveals how they are oriented in relation to each other. It can be calculated using the dot product and magnitudes of the vectors. The formula used to find the cosine of the angle (\(\theta\)) between two vectors {\(\mathbf{v}\)} and {\(\mathbf{w}\)} is:

$$ \mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos(\theta) $$

Rewriting the dot product equation in terms of \(\cos(\theta)\):

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

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

$$ \cos(\theta) = \frac{6}{(2\sqrt{2})(\sqrt{5})} = \frac{6}{2\sqrt{10}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10} $$

To find the angle \({\theta}\), take the inverse cosine:

$$ \theta = \cos^{-1}(\frac{3\sqrt{10}}{10}) $$

This angle is crucial for determining the relationship and orientation between the vectors.

Vector Properties

Knowing the properties of vectors helps in understanding their relationships. Vectors can be classified based on the angle between them.

• **Parallel Vectors**: If \(\theta\) is 0° or 180°, then \(\cos(\theta) = 1\) or \(\cos(\theta) = -1\), and the vectors are parallel.
• **Orthogonal Vectors**: If \(\theta\) is 90°, then \(\cos(\theta) = 0\), meaning the vectors are orthogonal (perpendicular).
• **Neither Parallel Nor Orthogonal**: Vectors do not fit the criteria for being parallel or orthogonal if the \(\cos(\theta)\) value is different from 1, -1, or 0.

Based on our example:

$$ \cos(\theta) = \frac{3\sqrt{10}}{10} $$

Since this value isn’t 1, -1, or 0, the vectors \({\mathbf{v}}\) and \({\mathbf{w}}\) are neither parallel nor orthogonal. Understanding these properties allows for more meaningful interpretations in vector analysis and applications.