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The dot product, sometimes referred to as the scalar product, is a fundamental operation in vector algebra. It combines two vectors to produce a scalar. This scalar is the sum of the products of the corresponding components of the vectors. The formula to compute the dot product of two three-dimensional vectors \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) and \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) is:\[\mathbf{v} \cdot \mathbf{u} = v_1u_1 + v_2u_2 + v_3u_3\]This operation is particularly useful for determining whether two vectors are orthogonal. If their dot product is zero, the vectors are perpendicular. In our example, the dot product of \( \mathbf{v} = 2 \mathbf{i} - 4 \mathbf{j} + \sqrt{5} \mathbf{k} \) and \( \mathbf{u} = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k} \) is calculated as follows:\[(2)(-2) + (-4)(4) + (\sqrt{5})(-\sqrt{5}) = -4 - 16 - 5 = -25\] Understanding the dot product helps in various applications like calculating work done by a force in physics.

The magnitude of a vector, often called its length or norm, indicates how long a vector is. It's denoted by \( |\mathbf{v}| \) and calculated using the Pythagorean theorem extended into three dimensions. For a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), its magnitude is given by:\[|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]This formula ensures that we're summing the squares of each component before taking the square root, a natural extension of calculating the hypotenuse of a right triangle. In our exercise, we have:- For \( \mathbf{v} = 2 \mathbf{i} - 4 \mathbf{j} + \sqrt{5} \mathbf{k} \),\[|\mathbf{v}| = \sqrt{2^2 + (-4)^2 + (\sqrt{5})^2} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5\]- Similarly, for \( \mathbf{u} = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k} \),\[|\mathbf{u}| = \sqrt{(-2)^2 + 4^2 + (-\sqrt{5})^2} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5\]Understanding magnitude is crucial in concepts like unit vectors and normalization, where vectors are scaled to a length of one.

Vector projection is a fascinating concept that involves projecting one vector onto another. The idea is to "shadow" one vector onto the direction of another, effectively simplifying a component of one vector relative to another. The formula for the projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is:\[\mathrm{proj}_\mathbf{v} \mathbf{u} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \mathbf{v}\]This formula scales the vector \( \mathbf{v} \) by the scalar obtained from the dot product divided by the square of the magnitude of \( \mathbf{v} \). It gives a vector with the same direction as \( \mathbf{v} \) but with a different magnitude. In the exercise, this calculation is done as follows:\[\mathrm{proj}_\mathbf{v} \mathbf{u} = \frac{-25}{25} \mathbf{v} = -1 \cdot (2 \mathbf{i} - 4 \mathbf{j} + \sqrt{5} \mathbf{k}) = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k}\]Projection is widely used in physics and engineering, especially in understanding forces, motion, and signals.

The angle between two vectors is a crucial measure that reveals how the vectors relate to one another in space. The cosine of this angle can be calculated using the dot product and the magnitudes of the vectors. For vectors \( \mathbf{v} \) and \( \mathbf{u} \), the cosine of the angle \( \theta \) is given by:\[\cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}||\mathbf{u}|}\]A cosine value can indicate whether the angle is acute, obtuse, or right:

• Positive cosine corresponds to an acute angle.
• Negative cosine indicates an obtuse angle.
• Zero signifies that the vectors are perpendicular.

In our problem, \( \cos \theta = \frac{-25}{25} = -1 \). This result implies that \( \theta \) is 180 degrees, meaning the vectors are in exactly opposite directions. Understanding the angle between vectors is not only fundamental in geometry but also in areas like computer graphics, where orientation and perspective are crucial.