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In vector algebra, the dot product is an important operation that helps us understand the relationship between two vectors. It is also known as the scalar product because it results in a single scalar quantity rather than another vector. The dot product of two vectors \(\overrightarrow{\mathbf{a}} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{b}} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}}\) is calculated as the sum of the products of their corresponding components:

• \( \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = a_1 \cdot b_1 + a_2 \cdot b_2 + a_3 \cdot b_3 \)

To better understand this concept, let's calculate the dot product of vectors \(\overrightarrow{\mathbf{a}} = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{b}} = 3 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}}\).

• The first component is \(2 \cdot 3 = 6\).
• The second component is \(1 \cdot (-4) = -4\).
• The third component is \(0 \cdot 0 = 0\).

Adding these, the dot product, \( \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = 6 - 4 + 0 = 2\). This dot product is particularly useful when calculating angles between vectors.

The magnitude of a vector, sometimes referred to as the length or norm, represents its size or length. It is a crucial concept in vector algebra, as it helps quantify how much of a quantity a vector represents. The magnitude of a vector \(\overrightarrow{\mathbf{a}} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}}\) is expressed as:

• \( \| \overrightarrow{\mathbf{a}} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

Let's find the magnitudes of the vectors used in the exercise. For \(\overrightarrow{\mathbf{a}} = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}\), the magnitude is computed as:

• \( \| \overrightarrow{\mathbf{a}} \| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \)

For \(\overrightarrow{\mathbf{b}} = 3 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}}\), the magnitude is:

• \( \| \overrightarrow{\mathbf{b}} \| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

Magnitude helps in comparing vectors and is indispensable when applying more complex operations such as calculating angles between vectors.

The cosine of the angle between two vectors is a crucial concept in vector algebra. It allows us to understand the directional relationship between vectors. The cosine formula relates the dot product of two vectors to their magnitudes and the cosine of the angle between them. For vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\), it is given by:

• \( \cos \theta = \frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{\| \overrightarrow{\mathbf{a}} \| \| \overrightarrow{\mathbf{b}} \|} \)

Applying this to the vectors in the given exercise, with the dot product of 2 and magnitudes \( \sqrt{6}\) and 5, we substitute into the formula:

• \( \cos \theta = \frac{2}{\sqrt{6} \times 5} = \frac{2}{5\sqrt{6}} \)
• To simplify, multiply both numerator and denominator by \(\sqrt{6}\) to rationalize the denominator: \( \frac{2\sqrt{6}}{30} = \frac{\sqrt{6}}{15} \)

This formula gives us a deep insight into how two vectors are aligned or misaligned in multi-dimensional space, making it easier to solve problems related to vector directions efficiently.