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When dealing with vectors in three-dimensional space, one important operation you can perform is the dot product. The dot product of two vectors results in a scalar—a single number—rather than another vector. For the vectors \( \mathbf{u} \) and \( \mathbf{v} \) given in the exercise, this is computed using their respective components.The formula to find the dot product is:

• If \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \), then \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \).

Using our vectors \( \mathbf{u} = (-1, 2, -3) \) and \( \mathbf{v} = (-1, -3, 4) \), we can substitute the components into the formula. This gives:\[ (-1)(-1) + (2)(-3) + (-3)(4) = 1 - 6 - 12 = -17 \].
This result, \(-17\), tells us something about the directions of the vectors relative to each other. In this case, since the dot product is negative, \( \mathbf{u} \) and \( \mathbf{v} \) are pointing away from each other in some sense.

The magnitude of a vector is a measure of its length. This is analogous to finding the length of a line segment in geometry and is particularly useful in understanding the size of a vector in three-dimensional vector calculus. To find the magnitude of a vector \( \mathbf{a} = (a_1, a_2, a_3) \), you use the formula:

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

Let's look at the magnitude of vectors \( \mathbf{u} \) and \( \mathbf{v} \):- For \( \mathbf{u} = (-1, 2, -3) \), calculate as follows: \[ \| \mathbf{u} \| = \sqrt{(-1)^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]- For \( \mathbf{v} = (-1, -3, 4) \), we compute: \[ \| \mathbf{v} \| = \sqrt{(-1)^2 + (-3)^2 + 4^2} = \sqrt{1 + 9 + 16} = \sqrt{26} \]These values of \( \sqrt{14} \) and \( \sqrt{26} \) give you the lengths of vectors \( \mathbf{u} \) and \( \mathbf{v} \) respectively. Knowing the magnitudes can be essential for calculating angles and determining vector norms in advanced applications.

Vectors are fundamental in representing quantities that have both a magnitude and a direction. In the realm of three-dimensional space, such vectors are expressed in the form \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \), where \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k} \) are standard unit vectors along the x, y, and z axes, respectively. This notation helps clearly designate the vector's orientation in 3D space.In our exercise, the vectors are:

• \( \mathbf{u} = -\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \)
• \( \mathbf{v} = -\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \)

Such notation demonstrates each vector's influence in three different dimensional directions. The components \((-1, 2, -3)\) for \( \mathbf{u} \) and \((-1, -3, 4)\) for \( \mathbf{v} \) show how each vector can be broken down into influences along the x, y, and z axes. This breakdown is critical for performing operations like addition, subtraction, and finding magnitudes or the dot product.By understanding three-dimensional vectors, you gain the ability to analyze the geometry and physics of more complex systems, from computer graphics to electromagnetic fields.