Problem 44 - Precalculus

Question

Find a unit vector in the direction of the given vector. $$\mathbf{v}=\langle-4 \sqrt{3},-2 \sqrt{3}\rangle$$

Step-by-Step Solution

Answer

The unit vector in the direction of \(\mathbf{v}\) is \(\left\langle -\frac{2\sqrt{3}}{\sqrt{15}}, -\frac{\sqrt{3}}{\sqrt{15}} \right\rangle\).

Step 1: Understanding the Goal

To find a unit vector in the direction of \(] = \langle -4 \sqrt{3}, -2 \sqrt{3} \rangle\), we need a vector in the same direction as \(\mathbf{v}\) that has a magnitude of 1.

Step 2: Calculate the Magnitude of the Original Vector

The first step in finding a unit vector is to calculate the magnitude of the given vector \(\mathbf{v}\). The formula for the magnitude of a vector \(\langle a, b \rangle\) is \(\sqrt{a^2 + b^2}\). For \(\mathbf{v} = \langle -4 \sqrt{3}, -2 \sqrt{3} \rangle\), we calculate: \[magnitude{]} = \sqrt{(-4 \sqrt{3})^2 + (-2 \sqrt{3})^2} = \sqrt{48 + 12} = \sqrt{60}\]\[magnitude{]} = \sqrt{60} = \sqrt{4 \cdot 15} = \sqrt{4} \cdot \sqrt{15} = 2 \sqrt{15}\]

Step 3: Normalize the Vector

To find the unit vector \(\mathbf{u}\), divide each component of \(\mathbf{v}\) by its magnitude. Thus, the unit vector is given by:\[\mathbf{u} = \frac{1}{2\sqrt{15}} \cdot \langle -4 \sqrt{3}, -2 \sqrt{3} \rangle = \left\langle \frac{-4 \sqrt{3}}{2 \sqrt{15}}, \frac{-2 \sqrt{3}}{2 \sqrt{15}} \right\rangle\]Each component simplifies to:\[\mathbf{u} = \left\langle \frac{-4}{2} \cdot \frac{\sqrt{3}}{\sqrt{15}}, \frac{-2}{2} \cdot \frac{\sqrt{3}}{\sqrt{15}} \right\rangle = \left\langle -2 \cdot \frac{\sqrt{3}}{\sqrt{15}}, -1 \cdot \frac{\sqrt{3}}{\sqrt{15}} \right\rangle\]Further simplifying gives:\[\mathbf{u} = \left\langle -\frac{2\sqrt{3}}{\sqrt{15}}, -\frac{\sqrt{3}}{\sqrt{15}} \right\rangle\]

Key Concepts

Vector Magnitude

When working with vectors, understanding vector magnitude is essential. Vector magnitude, often referred to as the "length" or "size" of a vector, measures how long or large the vector is in space. Given a vector \(\mathbf{v} = \langle a, b \rangle\), the formula to calculate its magnitude is \(\sqrt{a^2 + b^2}\). This formula comes from applying the Pythagorean theorem, which helps us find the hypotenuse length of a right triangle.

For example, with the vector \(\mathbf{v} = \langle -4\sqrt{3}, -2\sqrt{3} \rangle\), we calculate:

• Square each component: \((-4\sqrt{3})^2 = 48\) and \((-2\sqrt{3})^2 = 12\).
• Sum these squares: \(48 + 12 = 60\).
• Take the square root: \(\sqrt{60}\).

The magnitude can be further simplified using prime factorization: \(\sqrt{60} = \sqrt{4\cdot15} = 2\sqrt{15}\). Knowing the vector's magnitude is crucial for creating a unit vector.

Normalizing a Vector

Normalizing a vector converts any vector into a unit vector, which has a magnitude of exactly 1. This process involves dividing each component of the vector by its magnitude. This ensures the resulting vector retains the same direction as the original, but is scaled down to a single unit length.

For a given vector \(\mathbf{v}\) with magnitude \(magnitude_v\), the unit vector \(\mathbf{u}\) is found by:

• Dividing each component of \(\mathbf{v}\) by \(magnitude_v\)

Mathematically, if \(\mathbf{v} = \langle -4\sqrt{3}, -2\sqrt{3} \rangle\) and \(magnitude_v = 2\sqrt{15}\), then:

• \(\mathbf{u} = \langle \frac{-4\sqrt{3}}{2\sqrt{15}}, \frac{-2\sqrt{3}}{2\sqrt{15}} \rangle\)
• Simplifying, we get \(\mathbf{u} = \langle -2\frac{\sqrt{3}}{\sqrt{15}}, -\frac{\sqrt{3}}{\sqrt{15}} \rangle\)

Through normalization, the vector \(\mathbf{u}\) maintains its original direction but has a neatly set magnitude of 1.

Directional Vectors

Directional vectors are vectors that give us direction alone without regard to magnitude. By transforming a vector into a unit vector, it becomes a directional vector because its magnitude is precisely 1. This vector hence only indicates direction, making it particularly useful in applications where the direction is more vital than the length.

Such vectors are beneficial in:

• Pathfinding algorithms in computer science.
• Direction indicators in navigation systems.
• 3D modeling software to align objects correctly.

Using our previous vector \(\mathbf{v} = \langle -4\sqrt{3}, -2\sqrt{3} \rangle\) and finding \(\mathbf{u} = \langle -\frac{2\sqrt{3}}{\sqrt{15}}, -\frac{\sqrt{3}}{\sqrt{15}} \rangle\), converts the original vector to just a representation of direction.

Directional vectors are fundamental in scenarios where only orientation matters, ensuring consistently correct alignments and movements.