Problem 13 - Advanced Engineering Mathematics

Question

In Problems 13 and 14 , find \(\mathbf{a} \cdot \mathbf{b}\) if the smaller angle between a and \(\mathbf{b}\) is as given. $$ \|\mathbf{a}\|=10, \quad\|\mathbf{b}\|=5, \quad \theta=\pi / 4 $$

Step-by-Step Solution

Answer

The dot product \( \mathbf{a} \cdot \mathbf{b} = 25\sqrt{2} \).

Step 1: Understand the Problem

We need to find the dot product \( \mathbf{a} \cdot \mathbf{b} \) given the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \), and the angle between them. The given values are \( \|\mathbf{a}\| = 10 \), \( \|\mathbf{b}\| = 5 \), and the angle \( \theta = \frac{\pi}{4} \).

Step 2: Use the Dot Product Formula

The dot product \( \mathbf{a} \cdot \mathbf{b} \) can be calculated using the formula: \[ \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos \theta \] Substitute the known values into this formula.

Step 3: Calculate the Cosine of the Angle

The cosine of \( \theta = \frac{\pi}{4} \) is \( \cos \theta = \frac{\sqrt{2}}{2} \). We will use this value in our calculation.

Step 4: Substitute Values into the Formula

Substitute \( \|\mathbf{a}\| = 10 \), \( \|\mathbf{b}\| = 5 \), and \( \cos \theta = \frac{\sqrt{2}}{2} \) into the dot product formula: \[ \mathbf{a} \cdot \mathbf{b} = 10 \times 5 \times \frac{\sqrt{2}}{2} \]

Step 5: Perform the Multiplication

Calculate the product: \[ 10 \times 5 = 50 \]Then multiply by \( \frac{\sqrt{2}}{2} \): \[ 50 \times \frac{\sqrt{2}}{2} = 25\sqrt{2} \].

Key Concepts

Vector Magnitude

To get a clear understanding of vector magnitude, imagine a vector as an arrow in the space.It has both direction and length, but here, we focus on its length, known as the magnitude.The magnitude of a vector \(\mathbf{a}\), denoted as \(\|\mathbf{a}\|\), measures how long the arrow is.
For example, in our exercise, the given magnitude \(\|\mathbf{a}\| = 10\) means that the length of vector \(\mathbf{a}\) is 10 units. Similarly, \(\|\mathbf{b}\| = 5\) means vector \(\mathbf{b}\) is 5 units long.

• The magnitude is always a non-negative number.
• It provides useful information for various vector operations.

Understanding magnitude is crucial because it helps weigh how much each vector contributes to calculations like the dot product, directly impacting the results.
It's essentially the "size" of the vector in the vector space.

Angle Between Vectors

When dealing with vectors, knowing the angle between them is vital for determining how they interact.This angle, denoted as \( \theta \), can range from 0 to \(\pi\) (180 degrees) and illustrates their relative direction.
Here, it forms a central part of calculating the dot product.In the current exercise, \( \theta = \frac{\pi}{4} \) or 45 degrees. This specific angle indicates that the vectors are neither completely aligned (0 degrees) nor completely opposite (180 degrees).

• If \(\theta\) is 0, the vectors point in the same direction, maximizing their dot product.
• If \(\theta\) is 90 degrees, they are perpendicular, resulting in a dot product of zero.

Knowing \( \theta \) allows you to apply it in trigonometric functions to gain deeper insights into the vector relationship, as seen when calculating the dot product.

Trigonometric Functions

In vector analysis, trigonometric functions like cosine, sine, and tangent play a pivotal role.Among them, the cosine function is often used to find the dot product, which measures how much one vector projects onto another.
For the exercise in question, we calculate the cosine of the angle \( \theta = \frac{\pi}{4} \).The value \( \cos \theta = \frac{\sqrt{2}}{2} \) signifies the extent to which the vectors are aligned.

• Cosine values close to 1 indicate strong alignment.
• Values closer to 0 indicate perpendicular vectors.

By substituting this cosine value into the dot product formula, the interaction strength between vectors \(\mathbf{a}\) and \(\mathbf{b}\) becomes apparent.
This understanding of trigonometric functions not only guides the calculation but also illustrates how vectors relate spatially.