Problem 1 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-8: Find (a) the dot product of the two vectors and (b) the angle between the two vectors. $$ \langle-2,5\rangle, \quad\langle 3,6\rangle $$

Step-by-Step Solution

Answer

Dot product: 24; Angle: \( \theta = \cos^{-1} \left(\frac{8}{\sqrt{145}}\right) \).

Step 1: Understand the Vectors

Identify the given vectors as \( \mathbf{a} = \langle -2, 5 \rangle \) and \( \mathbf{b} = \langle 3, 6 \rangle \). We will find the dot product and the angle between these vectors.

Step 2: Calculate the Dot Product

The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated using the formula:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]Substitute the coordinates of the vectors:\[ \mathbf{a} \cdot \mathbf{b} = (-2)(3) + (5)(6) = -6 + 30 = 24 \]

Step 3: Find the Magnitude of the Vectors

Calculate the magnitude of each given vector. For \( \mathbf{a} = \langle -2, 5 \rangle \), the magnitude \( \| \mathbf{a} \| \) is:\[ \| \mathbf{a} \| = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \]For \( \mathbf{b} = \langle 3, 6 \rangle \), the magnitude \( \| \mathbf{b} \| \) is:\[ \| \mathbf{b} \| = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \]

Step 4: Use the Dot Product to Find the Angle

The cosine of the angle \( \theta \) between two vectors can be found using:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \]Substitute the values:\[ \cos \theta = \frac{24}{\sqrt{29} \times 3\sqrt{5}} = \frac{24}{3\sqrt{145}} = \frac{8}{\sqrt{145}} \]Therefore, the angle \( \theta \) can be calculated using \( \cos^{-1} \left(\frac{8}{\sqrt{145}}\right) \).

Key Concepts

Dot Product

The dot product is a way to multiply two vectors, which gives us a scalar (a single number) rather than another vector. This operation is particularly useful in various physics and engineering applications, as it helps determine quantities like work and similarity in directions. The formula for calculating the dot product of two vectors, say \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), is as follows:

\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]

For the vectors \( \mathbf{a} = \langle -2, 5 \rangle \) and \( \mathbf{b} = \langle 3, 6 \rangle \), we substitute the corresponding components:

• \( a_1 = -2 \), \( b_1 = 3 \)
• \( a_2 = 5 \), \( b_2 = 6 \)

Thus, the dot product is calculated as:
\[ \mathbf{a} \cdot \mathbf{b} = (-2)(3) + (5)(6) = -6 + 30 = 24 \]

The result of 24 tells us that there is a component of one vector in the direction of another, which can be interpreted in various physical contexts.

Magnitude of a Vector

The magnitude of a vector represents its length or size. When we look at vectors in a 2D plane, the magnitude can be compared to the distance from the origin to a point defined by the vector components. To calculate the magnitude of a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), we use the following formula:

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

Let's determine the magnitude of both vectors from the exercise. For \( \mathbf{a} = \langle -2, 5 \rangle \), the magnitude is:
\[ \| \mathbf{a} \| = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \]

Similarly, for \( \mathbf{b} = \langle 3, 6 \rangle \), the magnitude is:
\[ \| \mathbf{b} \| = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \]

These magnitudes are essential as they are used not only in calculating distances but also in determining the direction and angle between vectors.

Angle Between Vectors

Finding the angle between two vectors is crucial in understanding their spatial relationship. Whether vectors are pointing in the same direction, opposing directions, or are perpendicular can be vital information in many fields such as physics and computer graphics. The angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated using the dot product and the magnitudes of the vectors. The formula used is:

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

Substituting the known values from our exercise, we have:

• Dot product: \( 24 \)
• \( \| \mathbf{a} \| = \sqrt{29} \)
• \( \| \mathbf{b} \| = 3\sqrt{5} \)

This gives:
\[ \cos \theta = \frac{24}{\sqrt{29} \times 3\sqrt{5}} = \frac{24}{3\sqrt{145}} = \frac{8}{\sqrt{145}} \]

To find the angle \( \theta \), we take the arccosine:
\( \theta = \cos^{-1} \left(\frac{8}{\sqrt{145}}\right) \)

This calculation provides the actual angle between the two vectors, offering precise information about how they relate in space.