Problem 45 - Precalculus with Limits

Question

Finding a Unit Vector In Exercises \(39-48,\) find a unit vector in the direction of the given vector. Verify that the result has a magnitude of 1 . $$\mathbf{w}=4 \mathbf{j}$$

Step-by-Step Solution

Answer

The unit vector in the direction of \(\mathbf{w}=4 \mathbf{j}\) is \(\mathbf{j}\).

Step 1: Find the magnitude of the given vector

The magnitude of the vector \(\mathbf{w}=4 \mathbf{j}\) is the absolute value of 4 which is simply \(|\mathbf{w}| = 4\).

Step 2: Normalizing the vector

A unit vector is found by normalizing the original vector, which means dividing the vector by its magnitude. In this case, the unit vector \(\mathbf{u}\) in the direction of \(\mathbf{w}\) is given by \(\mathbf{u} = \frac{\mathbf{w}}{|\mathbf{w}|} = \frac{1}{4} \mathbf{w} = \frac{1}{4} \cdot 4 \mathbf{j} = 1 \cdot \mathbf{j} = \mathbf{j}\).

Step 3: Verify the magnitude of the unit vector

Verify that the magnitude of the unit vector is 1, that is, \(|\mathbf{u}|= 1\). Indeed, since \(\mathbf{u} = \mathbf{j}\), its magnitude is 1, and this shows that the vector is a unit vector.

Key Concepts

Vector Normalization

Vector normalization is the process of converting any given vector into a unit vector, which is a vector with a magnitude of exactly 1 that points in the same direction as the original vector. To normalize a vector, you divide each of its components by its magnitude. For example, if you have a vector \( \mathbf{v} \) with components \( (v_x, v_y, v_z) \) and magnitude \( |\mathbf{v}| \) (which is calculated using the formula \( \sqrt{v_x^2 + v_y^2 + v_z^2} \) for a 3D vector), you would get the normalized unit vector \( \mathbf{u} \) by computing \( \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = (\frac{v_x}{|\mathbf{v}|}, \frac{v_y}{|\mathbf{v}|}, \frac{v_z}{|\mathbf{v}|})\). By normalizing a vector, you simplify many calculations in physics and engineering, as it makes it much easier to express the direction without concerning the magnitude.

For the vector at hand, \( \mathbf{w}=4 \mathbf{j} \), normalizing would involve dividing each component by the vector's magnitude. But since the only non-zero component here is \( 4j \), along the y-axis, we simply divide it by the magnitude (which is 4 in this case), thus obtaining the unit vector \( \mathbf{j} \).

Vector Magnitude

The vector magnitude is the length or norm of the vector. It indicates the vector's size. In geometry and physics, this concept is fundamental because it gives an idea of how far something will move in a specific direction, when the vector is representing a force or a displacement for example. To calculate the magnitude of a vector with components \( (x, y, z) \), use the formula \( |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \), which is derived from the Pythagorean theorem.

For vectors in the plane (two-dimensional), the formula reduces to \( |\mathbf{v}| = \sqrt{x^2 + y^2} \). In the context of our exercise, the given vector \( \mathbf{w} \) only has a component in one dimension, \( 4 \mathbf{j} \), so its magnitude is simply the absolute value of that component, which is 4. Knowing the magnitude is crucial for the normalization process, as it's used to scale the vector to a unit vector.

Directional Vectors

Directional vectors are vectors that express the direction of a quantity, unambiguously ignoring its magnitude. They are extremely useful in physics and engineering because they can represent quantities like velocity, force, and displacement, with the direction being the primary focus. These vectors are typically represented as arrows pointing in the direction of the quantity, with their magnitude (or arrow length) reflecting the strength of the quantity they represent.

When we talk about unit vectors, we are referring to the specific kind of directional vectors that have a magnitude of 1. They are often used as the basis of a coordinate system; for example, the standard unit vectors in a 3D Cartesian coordinate system are \( \mathbf{i} \) (for the x-axis), \( \mathbf{j} \) (for the y-axis), and \( \mathbf{k} \) (for the z-axis). Our exercise deals with a vector that only involves the \( \mathbf{j} \) unit vector, meaning the direction is purely along the y-axis — an example of a directional vector in a real-world application might be gravity which, close to the Earth's surface, has a direction represented by a unit vector pointing downwards.