Problem 36 - Precalculus

Question

Determine whether each pair of vectors is orthogonal. $$\left\langle\frac{5}{6}, \frac{6}{7}\right\rangle \text { and }\left\langle\frac{36}{25},-\frac{49}{36}\right\rangle$$

Step-by-Step Solution

Answer

The vectors are not orthogonal because their dot product is not zero.

Step 1: Recall the condition for orthogonality

Two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) are orthogonal if their dot product is equal to zero. The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as \( a_1b_1 + a_2b_2 \).

Step 2: Assign values to vector components

Identify the components of the vectors provided: \( \mathbf{a} = \left\langle \frac{5}{6}, \frac{6}{7} \right\rangle \) and \( \mathbf{b} = \left\langle \frac{36}{25}, -\frac{49}{36} \right\rangle \). Here, \( a_1 = \frac{5}{6} \), \( a_2 = \frac{6}{7} \), \( b_1 = \frac{36}{25} \), and \( b_2 = -\frac{49}{36} \).

Step 3: Calculate the dot product

Substitute the components into the dot product formula: \( \mathbf{a} \cdot \mathbf{b} = \left(\frac{5}{6} \cdot \frac{36}{25}\right) + \left(\frac{6}{7} \cdot -\frac{49}{36}\right) \).

Step 4: Simplify the dot product terms

Calculate each part of the dot product separately. Start with the first term: \( \frac{5}{6} \cdot \frac{36}{25} = \frac{180}{150} = \frac{6}{5} \). Now, the second term: \( \frac{6}{7} \cdot -\frac{49}{36} = -\frac{294}{252} = -\frac{7}{6} \).

Step 5: Check if dot product equals zero

Add the simplified terms: \( \frac{6}{5} + -\frac{7}{6} = \frac{36}{30} - \frac{35}{30} = \frac{1}{30} \). Since \( \frac{1}{30} eq 0 \), the vectors are not orthogonal.

Key Concepts

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers, usually represented as vectors, and returns a single number. This operation is fundamental in determining whether two vectors are orthogonal. Let's break it down:

• The dot product is calculated by multiplying corresponding components of the vectors and summing these products.
• For example, if we have two vectors, \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is formally expressed as \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).
• The result is a scalar value, which can tell us a lot about the relationship between the vectors.

In the example given, substituting the components of the vectors into this formula gives a result that helps us determine their relationship.

Vector Components

When discussing vectors, we often break them down into their components, which are their individual parts in each dimension. These are essential when performing operations like the dot product.

• A vector in two dimensions is represented as \( \langle x, y \rangle \), where \( x \) and \( y \) are the components.
• For example, vector \( \mathbf{a} = \langle \frac{5}{6}, \frac{6}{7} \rangle \) has components \( a_1 = \frac{5}{6} \) and \( a_2 = \frac{6}{7} \).
• Similarly, vector \( \mathbf{b} = \langle \frac{36}{25}, -\frac{49}{36} \rangle \) has components \( b_1 = \frac{36}{25} \) and \( b_2 = -\frac{49}{36} \).

Understanding the components is key when substituting them into the formula for the dot product. Each component pairs up with its counterpart to contribute to the product calculation. This clarity can help in solving various vector-related problems.

Orthogonality Condition

The concept of orthogonality in vectors is equivalent to perpendicularity in geometry. Two vectors are said to be orthogonal if their dot product is zero.

• Mathematically, if \( \mathbf{a} \cdot \mathbf{b} = 0 \), then vectors \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal.
• This condition indicates that the vectors do not "lean" towards or against each other, i.e., they are independent in terms of direction.
• In our example, after computing the dot product we found \( \frac{1}{30} \), which is not zero, thus confirming that the vectors are not orthogonal.

Orthogonality is particularly important in many fields like physics and engineering, where it can simplify the analysis of forces, movements, and signals. Understanding this concept enables making sense of complex systems by analyzing components separately.