Problem 63 - 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. $$ |\mathbf{u}|=10,|\mathbf{v}|=5, \theta=\pi / 4 $$

Step-by-Step Solution

Answer

The dot product \( \mathbf{u} \cdot \mathbf{v} = 25\sqrt{2} \).

Step 1: Recall the Dot Product Formula

The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by the formula: \[ \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos(\theta) \] where \( \theta \) is the angle between the two vectors.

Step 2: Plug in Given Values

Substitute the given values into the dot product formula. You have \( |\mathbf{u}| = 10 \), \(|\mathbf{v}| = 5\), and \( \theta = \pi/4 \). So the expression becomes: \[ \mathbf{u} \cdot \mathbf{v} = 10 \times 5 \times \cos\left(\frac{\pi}{4}\right) \]

Step 3: Calculate \(\cos(\pi/4)\)

The cosine of \(\pi/4\) is \(\frac{\sqrt{2}}{2}\). Substitute this value into the equation: \[ \mathbf{u} \cdot \mathbf{v} = 10 \times 5 \times \frac{\sqrt{2}}{2} \]

Step 4: Simplify the Expression

Now perform the multiplication: \[ 10 \times 5 = 50 \] and \[ 50 \times \frac{\sqrt{2}}{2} = 25\sqrt{2} \]. This is the value of the dot product.

Key Concepts

Vectors

Vectors are fundamental tools in mathematics and physics used to represent quantities that have both magnitude and direction. Think of a vector as an arrow pointing from one place to another, such as a force pointing upward or a velocity moving eastward.

• **Components**: Vectors can be broken down into components, usually along x and y-axes in a 2D plane. For example, a vector "3i + 4j" has components 3 and 4 along the x and y directions, respectively.
• **Notation**: Vectors are often written in bold, like \( \mathbf{u} \) or \( \mathbf{v} \), or with an arrow on top, such as \( \vec{u} \).
• **Operations**: Key operations involving vectors include addition, subtraction, and multiplication (dot product and cross product).

Understanding vectors is crucial when dealing with mathematical problems involving direction as well as magnitude.

Cosine of Angle

The cosine of an angle is a trigonometric function that relates the angle to the ratio of the length of the adjacent side to the hypotenuse in a right-angled triangle. When dealing with vectors, the cosine of the angle helps determine how two vectors relate to each other spatially.

• **Range**: The cosine of an angle can vary from -1 to 1. When \(\cos\theta = 1\), vectors are pointing in the same direction.
• **Usage in Dot Product**: In the formula for the dot product \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos(\theta) \), it shows how much of one vector goes in the direction of the other. The cosine factor determines the influence based on the angle.

For example, in the exercise provided, \(\theta = \frac{\pi}{4}\) radians is a common angle where \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \). This value is crucial for calculating the dot product.

Vector Magnitude

The magnitude of a vector is a measure of its length or size, often calculated as the distance between its origin and endpoint in space.

• **Calculation**: For a vector \( \mathbf{a} = \langle a_x, a_y \rangle \), the magnitude \( |\mathbf{a}| \) is calculated using the formula \( \sqrt{a_x^2 + a_y^2} \) in two dimensions.
• **Application**: Magnitude is used to determine how "strong" or "forceful" a vector is. It's a key factor in the dot product, affecting the product's scale directly.

In our exercise example, the magnitudes of vectors \( \mathbf{u} \) and \( \mathbf{v} \) were given as 10 and 5. These values were crucial inputs for calculating the dot product, highlighting how essential magnitude is in vector mathematics.