Problem 64 - Precalculus with Calculus Previews

Question

Find the dot product \(\mathbf{u} \cdot \mathbf{v}\) if the smaller angle between \(\mathbf{u}\) and \(\mathbf{v}\) is as given. $$ |\mathbf{u}|=6,|\mathbf{v}|=12, \theta=\pi / 6 $$

Step-by-Step Solution

Answer

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is \( 36\sqrt{3} \).

Step 1: Understanding the Dot Product Formula

To find the dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) when the angle \( \theta \) between them is known, we use the formula: \( \mathbf{u} \cdot \mathbf{v} = |{u}| |{v}| \cos(\theta) \). This formula involves the magnitudes of the vectors and the cosine of the angle between them.

Step 2: Identifying Known Values

We are given \( |{u}| = 6 \), \( |{v}| = 12 \), and \( \theta = \frac{\pi}{6} \). These values will be used in the dot product formula.

Step 3: Using the Cosine Value

Calculate \( \cos(\frac{\pi}{6}) \). Knowing that \( \frac{\pi}{6} \) is 30 degrees, we use the cosine value \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \).

Step 4: Calculating the Dot Product

Insert the known values into the dot product formula: \( \mathbf{u} \cdot \mathbf{v} = 6 \times 12 \times \frac{\sqrt{3}}{2} \).

Step 5: Simplifying the Expression

Perform the multiplication: \( 6 \times 12 = 72 \). Then, \( 72 \times \frac{{3}}{2} \equiv 36\sqrt{3} \). Therefore, \( \mathbf{u} \cdot \mathbf{v} = 36\sqrt{3} \).

Key Concepts

Vector Magnitude

In the world of vectors, the magnitude, also known as the length or the norm, represents the size of a vector. Imagine it like the distance from the origin to the point where the vector ends in a coordinate system. It quantifies how "long" the vector is. Calculating the magnitude is straightforward when given the vector components, or in this case, it is provided directly. For vector \( \mathbf{u} \), the magnitude is denoted as \( |\mathbf{u}| \) and for vector \( \mathbf{v} \), it is \( |\mathbf{v}| \).
To apply vector magnitude effectively in problems like the dot product, remember:

• The magnitude is always a non-negative number.
• It is crucial for calculating distances and understanding the space a vector spans.
• In this exercise, \( |\mathbf{u}| = 6 \) and \( |\mathbf{v}| = 12 \), showcasing their lengths.

Understanding vector magnitudes allows us to proceed with calculating properties like the dot product, that reveal relationships between vectors.

Angle Between Vectors

The angle between vectors is a fundamental concept, especially when dealing with their interactions. This angle, denoted as \( \theta \), is the measure of rotation needed for one vector to align with another. It plays a significant role in defining the orientation relationship between vectors.
The cosine of this angle is what you use when finding the dot product. In this specific problem, \( \theta = \frac{\pi}{6} \), which corresponds to 30 degrees. A smaller angle, like this one, indicates the vectors are relatively aligned and not perpendicular or opposite.
This angle determines how much of one vector goes in the direction of the other. When the angle is smaller, the vectors have more in common directionally. Choosing angles like \( \frac{\pi}{6} \), which are common in trigonometry, simplifies calculations. The consistent use of this value helps in memorizing trigonometric functions, supporting further problem-solving.

Cosine Function

The cosine function is vital in calculating the dot product of vectors when their magnitude and the angle between them are known. It relates to the x-coordinate on the unit circle for a given angle. Often used in geometry and physics, the cosine function emphasizes the importance of the angular relationship between vectors.
For \( \theta = \frac{\pi}{6} \), the cosine value is \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \). This value is derived from trigonometric identities or the unit circle. Learning these values helps to quickly compute problems involving angles.

• \( \cos(0^\circ) = 1 \).
• \( \cos(90^\circ) = 0 \).
• For acute angles often used, like \( 30^\circ \), the cosine function outputs provide easy multiplication with vector magnitudes.

Understanding and applying the cosine function is crucial in transforming angular information into scalar products, like the dot product, where the cosine modifies the sizes of vectors proportionally to how much they are aligned.