Problem 21 - Advanced Engineering Mathematics

Question

In Problems 21-24, find the angle \(\theta\) between the given vectors. $$ \mathbf{a}=3 \mathbf{i}-\mathbf{k}, \mathbf{b}=2 \mathbf{i}+2 \mathbf{k} $$

Step-by-Step Solution

Answer

The angle \( \theta \) is approximately \( 63.43^{\circ} \).

Step 1: Find Dot Product

First, find the dot product of the vectors \( \mathbf{a} \) and \( \mathbf{b} \). The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated using the formula: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). Here, \( \mathbf{a} = 3 \mathbf{i} + 0 \mathbf{j} - \mathbf{k} \) and \( \mathbf{b} = 2 \mathbf{i} + 0 \mathbf{j} + 2 \mathbf{k} \). So, calculate: \[ \mathbf{a} \cdot \mathbf{b} = (3)(2) + (0)(0) + (-1)(2) = 6 + 0 - 2 = 4. \]

Step 2: Find Magnitude of Vectors

Calculate the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \) using the formula: \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \). For vector \( \mathbf{a} = 3\mathbf{i} - \mathbf{k} \), magnitude \( \| \mathbf{a} \| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}. \) Similarly, for \( \mathbf{b} = 2\mathbf{i} + 2\mathbf{k} \), \( \| \mathbf{b} \| = \sqrt{2^2 + 0^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}. \)

Step 3: Calculate Cosine of Angle

Use the dot product and magnitudes to find \( \cos \theta \) from the relationship: \( \mathbf{a} \cdot \mathbf{b} = \| \mathbf{a} \| \| \mathbf{b} \| \cos \theta \). Substitute the known values into the formula: \[ 4 = \sqrt{10} \cdot 2\sqrt{2} \cdot \cos \theta. \] Thus, \[ \cos \theta = \frac{4}{2\sqrt{20}} = \frac{4}{4\sqrt{5}} = \frac{1}{\sqrt{5}}. \]

Step 4: Find the Angle \( \theta \)

Find the angle \( \theta \) using the inverse cosine function: \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \). Use a calculator to estimate \( \theta \approx 63.43^{\circ} \).

Key Concepts

Dot Product

In vector algebra, understanding the dot product is essential. It is a way to multiply two vectors, resulting in a scalar quantity. The dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) is computed using their components:

• \( \mathbf{a} \cdot \mathbf{b} = a_1 \times b_1 + a_2 \times b_2 + a_3 \times b_3 \)

This formula efficiently calculates the dot product by summing the products of corresponding components. It's handy for assessing the vectors' directional alignment.
A positive dot product suggests that vectors point in the same general direction, while a negative result indicates diverging directions. In our example, vectors \( \mathbf{a} = 3 \mathbf{i} - \mathbf{k} \) and \( \mathbf{b} = 2 \mathbf{i} + 2 \mathbf{k} \) have their dot product calculated as:

• \[ \mathbf{a} \cdot \mathbf{b} = (3)(2) + (0)(0) + (-1)(2) = 6 + 0 - 2 = 4 \]

This indicates a tendency towards the same direction, yet not fully aligned.

Magnitude of a Vector

The magnitude of a vector is its length or size. You can think of it as the distance of the vector from the origin in a 3D space. Calculating the magnitude of a vector \( \mathbf{v} \) involves its components:

• \[ \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]

Each component is squared, summed up, and then the square root is taken. This gives us the Euclidean length of the vector.
For example, the magnitude of vector \( \mathbf{a} = 3\mathbf{i} - \mathbf{k} \) is:

• \[ \| \mathbf{a} \| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

Similarly, for vector \( \mathbf{b} = 2\mathbf{i} + 2\mathbf{k} \), the magnitude is:

• \[ \| \mathbf{b} \| = \sqrt{2^2 + 0^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

Understanding magnitudes is crucial when creating comparisons or calculating angles in vector problems.

Cosine of Angle Between Vectors

Once you have both the dot product and magnitudes of vectors, finding the cosine of the angle \( \theta \) between them becomes straightforward. The formula used is:

• \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \]

This relation stems from the geometric definition of the dot product. Solving for \( \theta \) ensures understanding of how aligned two vectors are.
In our case, the values are plugged into:

• \[ \cos \theta = \frac{4}{\sqrt{10} \cdot 2\sqrt{2}} = \frac{1}{\sqrt{5}} \]

Finally, to find the angle \( \theta \), you apply the inverse cosine function:

• \[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \]

Using a calculator, you approximate \( \theta \approx 63.43^\circ \). This calculation allows you to visualize how much one vector has to rotate to align with another.