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When discussing vectors, their magnitude is a fundamental concept. The magnitude of a vector measures its length. For a vector represented as \(\mathbf{u} = \langle x, y \rangle\), its magnitude \(\|\mathbf{u}\|\) is calculated using the Pythagorean theorem:

• Magnitude Formula: \(\|\mathbf{u}\| = \sqrt{x^2 + y^2}\)

In the given problem, the vector \(\mathbf{u} = \langle 2, -1 \rangle\) has a magnitude calculated as \(\sqrt{2^2 + (-1)^2} = \sqrt{5}\). Understanding how to find a vector's magnitude is critical, as it helps in normalizing vectors and finding vectors of a specific length.
The magnitude provides insight into the size of the vector, independent of its direction, allowing comparisons and transformations (such as creating unit vectors) easily.

A unit vector is a vector that has a magnitude of 1. It does not affect the direction, making it a handy reference for direction alone. Transforming any vector into a unit vector involves dividing the vector by its magnitude. Given a vector \(\mathbf{u} = \langle x, y \rangle\), its unit vector \(\mathbf{u_0}\) is calculated as:

• Unit Vector Formula: \(\mathbf{u_0} = \left(\frac{x}{\|\mathbf{u}\|}, \frac{y}{\|\mathbf{u}\|}\right)\)

In our exercise, with \(\mathbf{u} = \langle 2, -1 \rangle\) and \(\|\mathbf{u}\| = \sqrt{5}\), the unit vector \(\mathbf{u_0}\) becomes \(\left( \frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}} \right)\).
This unit vector exactly matches the direction of \(\mathbf{u}\) but scaled to a magnitude of 1. Unit vectors are crucial for applying transformations and scaling vectors to desired lengths, as they preserve direction integrity.

Scalar multiplication involves multiplying a vector by a scalar (a single number), which results in a new vector. This operation changes the magnitude of the vector but not its direction unless the scalar is negative, which reverses it. The process for scalar multiplication of a vector \(\mathbf{u} = \langle x, y \rangle\) by a scalar \k\ is:

• Scalar Multiplication Formula: \(\mathbf{v} = k \cdot \mathbf{u} = \langle kx, ky \rangle\)

In the problem, the unit vector \(\mathbf{u_0}\) was scaled by a scalar multiple of 10 to get vector \(\mathbf{v}\) with the desired magnitude of 10:

• Calculation: \(\mathbf{v} = 10 \cdot \left(\frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right) = \left(4\sqrt{5}, -2\sqrt{5}\right)\)

This vector \(\mathbf{v}\) maintains the same direction as \(\mathbf{u}\) and achieves the target magnitude when simplified. Understanding this concept allows one to manipulate vectors to represent different quantities or forces through scalar alteration.