Problem 22 - Advanced Engineering Mathematics

Question

Find the angle \(\theta\) between the given vectors. $$ \mathbf{a}=2 \mathbf{i}+\mathbf{j}, \mathbf{b}=-3 \mathbf{i}-4 \mathbf{j} $$

Step-by-Step Solution

Answer

The angle \(\theta\) is approximately \(116.57^\circ\).

Step 1: Calculate Dot Product of Vectors

To find the angle \(\theta\) between two vectors, we start with the dot product formula. The dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).For \( \mathbf{a}=2 \mathbf{i}+\mathbf{j} \), let \( a_1 = 2 \) and \( a_2 = 1 \).For \( \mathbf{b}=-3 \mathbf{i}-4 \mathbf{j} \), let \( b_1 = -3 \) and \( b_2 = -4 \).Thus, \( \mathbf{a} \cdot \mathbf{b} = (2)(-3) + (1)(-4) = -6 - 4 = -10 \).

Step 2: Calculate Magnitudes of Each Vector

Next, calculate the magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\). The magnitude of a vector \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}\) is calculated by the formula \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \).For \( \mathbf{a}=2 \mathbf{i}+\mathbf{j} \):\[ \|\mathbf{a}\| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]For \( \mathbf{b}=-3 \mathbf{i}-4 \mathbf{j} \):\[ \|\mathbf{b}\| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5 \]

Step 3: Use the Dot Product and Magnitudes to Find the Angle

Now that we have the dot product and magnitudes, apply the cosine formula.The cosine of the angle \(\theta\) between vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]Substituting the values we calculated:\[ \cos \theta = \frac{-10}{\sqrt{5} \times 5} = \frac{-10}{5\sqrt{5}} = \frac{-2}{\sqrt{5}} \]

Step 4: Solve for \(\theta\)

To find \(\theta\), we take the inverse cosine (arccos) of the result from Step 3.\[ \theta = \cos^{-1} \left( \frac{-2}{\sqrt{5}} \right) \]

Step 5: Calculate the Value of \(\theta\)

Using a calculator to evaluate \( \theta = \cos^{-1} \left( \frac{-2}{\sqrt{5}} \right) \), we find the angle in degrees or radians as needed. Numerically, this gives \( \theta \approx 116.57^\circ \) when evaluated in degrees.

Key Concepts

Dot Product

The dot product is a way of multiplying two vectors to get a scalar, which contains information about the magnitude and the direction of the vectors. When you calculate the dot product, you're essentially measuring how much one vector "goes" in the direction of another.
A vector is often written in terms of its components, like \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \). The dot product formula between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In simpler terms, you multiply the corresponding components of the two vectors and sum the results.

In the exercise, one vector was \( \mathbf{a} = 2 \mathbf{i} + \mathbf{j} \) with components \( a_1 = 2 \) and \( a_2 = 1 \). The other vector was \( \mathbf{b} = -3 \mathbf{i} - 4 \mathbf{j} \) with components \( b_1 = -3 \) and \( b_2 = -4 \).

By substituting these values into the dot product formula, you get:

• \( \mathbf{a} \cdot \mathbf{b} = 2(-3) + 1(-4) = -10 \)

Understanding this calculation is crucial because the dot product helps us find the angle between the vectors, which reveals how aligned they are.

Magnitude of a Vector

The magnitude of a vector, often represented by \( \|\mathbf{a}\| \), is essentially the "length" or "size" of the vector. It tells us how far a vector stretches from its starting point. Calculating the magnitude involves using the Pythagorean theorem in the context of a vector's components.

For a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \), the magnitude is given by the formula:

• \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \)

This means you square each component of the vector, sum them, and then take the square root.

In the specific problem, the vector \( \mathbf{a} = 2\mathbf{i} + \mathbf{j} \) has components \( a_1 = 2 \) and \( a_2 = 1 \). Therefore, the magnitude is:

• \( \|\mathbf{a}\| = \sqrt{2^2 + 1^2} = \sqrt{5} \)

The second vector \( \mathbf{b} = -3\mathbf{i} - 4\mathbf{j} \) has a magnitude of:

• \( \|\mathbf{b}\| = \sqrt{(-3)^2 + (-4)^2} = 5 \)

Getting these magnitudes is necessary for evaluating how vectors compare in size and position.

Angle Between Vectors

To find the angle between vectors, we utilize both the dot product and the magnitudes. The angle \( \theta \) can be found using the cosine formula which relates these quantities.

The formula for finding the cosine of the angle between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

This formula essentially combines the dot product, which measures how much two vectors "agree" in direction, and the magnitudes, which tell us their lengths.

Using the calculated dot product \( -10 \) and the magnitudes \( \sqrt{5} \) and \( 5 \), we find \( \cos \theta \):

• \( \cos \theta = \frac{-10}{\sqrt{5} \times 5} = \frac{-2}{\sqrt{5}} \)

Once we have \( \cos \theta \), we can solve for \( \theta \) by determining the inverse cosine (arccos) of \( \frac{-2}{\sqrt{5}} \). This yields the angle between the vectors.

In this case, \( \theta \approx 116.57^\circ \), indicating how the vectors diverge from each other. Understanding the geometric relationship between vectors is important in vector calculus and helps solve many physical and engineering problems.