Problem 40 - 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 . $$\mathbf{u}=\langle 0,-2\rangle$$

Step-by-Step Solution

Answer

The unit vector in the direction of the given vector \( \mathbf{u}=\langle 0,-2\rangle \) is \( \langle 0, -1 \rangle \)

Step 1: Find the Magnitude of the Given Vector

The formula to find the magnitude of a vector \( \mathbf{u}=\langle a,b\rangle \) is \( ||\mathbf{u}|| = \sqrt{a^2 + b^2} \). Now apply this formula to calculate the magnitude of the given vector \( \mathbf{u}=\langle 0,-2\rangle \), \( ||\mathbf{u}|| = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2 \).

Step 2: Find Unit Vector

The unit vector in the direction of a given vector is obtained by dividing the original vector by its magnitude, so the unit vector \( u_{unit} = \frac{\mathbf{u}}{||\mathbf{u}||} = \frac{\langle 0, -2\rangle}{2} = \langle 0, -1 \rangle \).

Step 3: Verify the Magnitude of the Unit Vector

To confirm this result, find the magnitude of the unit vector and verify that it is 1 . Use the formula \( ||\mathbf{u}|| = \sqrt{a^2 + b^2} \) again, \( ||u_{unit}|| = \sqrt{(0)^2 + (-1)^2} = \sqrt{1} = 1 \). As expected the magnitude of the unit vector is 1.

Key Concepts

Vector Magnitude

Every vector has a magnitude, which tells you how long or strong that vector is. You can think of it like the length of a line stretching from the vector’s starting point (often the origin) to its endpoint.
The formula to find the magnitude of a two-dimensional vector \( \mathbf{u} = \langle a, b \rangle \) is \( ||\mathbf{u}|| = \sqrt{a^2 + b^2} \). This formula works similarly to the Pythagorean Theorem, where the vector components \(a\) and \(b\) form a right triangle.

• For example, for the vector \( \mathbf{u} = \langle 0, -2 \rangle \), the magnitude is calculated as \( ||\mathbf{u}|| = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2 \).

Knowing the magnitude is essential when you want to work with unit vectors, as it helps you to "normalize" the given vector by shrinking or stretching it to a length of one.

Vector Direction

The direction of a vector is the path that the vector moves towards in its space, resembling the arrow’s tip pointing from the vector’s origin to its end.
Vector direction is generally expressed using directional angles or by simply considering the vector components.

• When the direction is provided as an angle, trigonometric functions like sine, cosine, and tangent can be used to interpret the precise direction.
• In simple cases, like \( \mathbf{u} = \langle 0, -2 \rangle \), you may describe the direction based on its components, noticing in this example that it moves entirely downward on the vertical axis.

Understanding a vector's direction is crucial when converting it to a unit vector, as the direction remains unchanged during the conversion.

Vector Normalization

Normalization is the process of converting a vector into a unit vector. A unit vector has a magnitude of 1 but retains the original direction of the vector. This concept is vital in mathematics and physics, especially in applications involving direction without concern for size.
To normalize a vector, divide each component of the vector by its magnitude:

• If the vector \( \mathbf{u} = \langle a, b \rangle \) has a magnitude \( ||\mathbf{u}|| \), the unit vector \( \mathbf{u}_{unit} \) is found as \( \mathbf{u}_{unit} = \frac{1}{||\mathbf{u}||} \langle a, b \rangle \).
• For our example vector \( \mathbf{u} = \langle 0, -2 \rangle \), with a magnitude of 2, the normalized vector is \( \langle 0, -1 \rangle \): \( \mathbf{u}_{unit} = \frac{\langle 0, -2 \rangle}{2} \).
• Verify by checking the magnitude of the resulting unit vector: it should be 1.

By using vector normalization, different vectors can be compared on a common scale, which is fundamental in various computational and geometric applications.