Problem 8 - 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. $$ \langle-3,6\rangle, \quad\langle-1,2\rangle $$

Step-by-Step Solution

Answer

(a) Dot product: 15; (b) Angle: 0 degrees.

Step 1: Understanding the Dot Product Formula

The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) in two-dimensional space is given by the formula: \( \mathbf{a} \cdot \mathbf{b} = a_1 \times b_1 + a_2 \times b_2 \). This formula involves multiplying the corresponding components of the vectors and adding the results.

Step 2: Apply the Dot Product Formula

For vectors \( \langle -3, 6 \rangle \) and \( \langle -1, 2 \rangle \), assign: \( a_1 = -3 \), \( a_2 = 6 \), \( b_1 = -1 \), \( b_2 = 2 \). Compute: \[ \mathbf{a} \cdot \mathbf{b} = (-3)(-1) + (6)(2) = 3 + 12 = 15 \] The dot product is 15.

Step 3: Calculating the Magnitudes of the Vectors

The magnitude of a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \) is calculated using \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \). For \( \mathbf{a} = \langle -3, 6 \rangle \): \[ \|\mathbf{a}\| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \] For \( \mathbf{b} = \langle -1, 2 \rangle \): \[ \|\mathbf{b}\| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

Step 4: Use the Dot Product to Find the Angle

To find the angle \( \theta \) between two vectors, use the formula: \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \). Substitute the dot product and magnitudes: \[ \cos \theta = \frac{15}{(3\sqrt{5})(\sqrt{5})} = \frac{15}{15} = 1 \] Since \( \cos \theta = 1 \), the angle \( \theta = \arccos(1) = 0 \) degrees.

Key Concepts

Vectors

Vectors are essential components in mathematics and physics, representing quantities that have both a magnitude and a direction. One can think of a vector as an arrow pointing in a certain direction, where:* The length of the arrow represents the magnitude, and * The direction in which the arrow points shows the vector's direction.
When writing vectors in mathematical terms, we commonly use angled brackets with components, such as \( \langle -3, 6 \rangle \). These components show how the vector moves within a space: * The first number represents movement along the x-axis, and * The second number represents movement along the y-axis.
In this exercise, the vectors \( \langle -3, 6 \rangle \) and \( \langle -1, 2 \rangle \) are given. The goal is to find mathematical properties like the dot product and the angle between these vectors.

Magnitude

The magnitude of a vector is like measuring the length of the vector's arrow. It shows how much "stuff" the vector carries, regardless of its direction. To calculate the magnitude of a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \) in two-dimensional space, use the formula: \[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \]
This formula involves: * Squaring each component of the vector, * Adding those squared values, and then * Taking the square root of the result.
For example, for vector \( \langle -3, 6 \rangle \), the magnitude becomes: \[ \|\mathbf{a}\| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = 3\sqrt{5} \]
This calculation provides the magnitude value, which is always a non-negative number.

Angle Between Vectors

Finding the angle between two vectors helps understand how they are oriented relative to each other in space. The angle \( \theta \) is the measure of separation between them. A common approach to find this angle is by using the dot product formula.
For vectors \( \mathbf{a} \) and \( \mathbf{b} \), with the dot product \( \mathbf{a} \cdot \mathbf{b} \) already calculated, the angle \( \theta \) can be found using: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]
This provides the cosine of the angle. To find the angle itself, you would take the inverse cosine (arccos).

• If \( \cos \theta = 1 \), the angle is \( 0 \) degrees (vectors point in the same direction).
• If \( \cos \theta = -1 \), the angle is \( 180 \) degrees (vectors point in opposite directions).
• If \( \cos \theta = 0 \), the angle is \( 90 \) degrees (vectors are perpendicular).

For this exercise, the computed \( \cos \theta = 1 \), indicating that the vectors are aligned and \( \theta = 0 \) degrees.

Cosine Formula

The cosine formula is prominently used to determine the angle between two vectors. This formula links the dot product of two vectors to their magnitudes and the cosine of the angle between them. Written as: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]
The numerator is the dot product of the vectors \( \mathbf{a} \) and \( \mathbf{b} \). This gives an overall measure that reflects how much one vector "overlaps" with the other in terms of direction. The denominator, a product of the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \), scales the relationship, allowing the formula to solve for \( \cos \theta \).
The cosine value itself informs about the relationship of the vectors:

• If \( \cos \theta \) is positive, vectors largely point in the same general direction.
• If it's negative, they are more in opposite directions.
• If zero, vectors are perfectly perpendicular to each other.

This comprehensible explanation of the formula is key to finding angles in vector mathematics and applies to both this specific exercise and broader vector analysis tasks.