Finding Constant K Given Vectors And Cosine Of Angle

Introduction

In the realm of vector algebra, understanding the relationships between vectors, particularly the angles they form, is crucial. This article delves into a problem where we are given two vectors, x and y, defined in terms of unit vectors i and j, and a constant k. Our task is to determine the possible values of k given that the cosine of the angle between x and y is ${\frac{\sqrt{5}}{5}}$. This involves applying concepts of dot products, vector magnitudes, and trigonometric relationships. The process will illustrate how these mathematical tools can be used to solve geometric problems in a vector space. By carefully working through the steps, we'll uncover the solution and solidify our understanding of vector algebra. Let's embark on this mathematical journey to unravel the mystery of the constant k.

Problem Statement

We are given two vectors:

• x = 3i - j
• y = 2i + kj*

We also know that the cosine of the angle between x and y is ${\frac{\sqrt{5}}{5}}$. Our objective is to find the possible values of the constant k. This problem combines vector algebra with trigonometry, requiring us to utilize the dot product formula and the relationship between dot product and the cosine of the angle between two vectors. The challenge lies in setting up the equations correctly and solving for k, which may involve algebraic manipulation and potentially lead to multiple solutions. Understanding vector components and their interactions is key to unlocking the solution to this problem. Let's break down the steps and tackle this mathematical puzzle.

Methodology

To solve this problem, we'll employ the following steps:

1. Calculate the dot product of vectors x and y: The dot product, denoted as x · y, can be calculated using the components of the vectors. For x = 3i - j and y = 2i + kj*, the dot product is (3)(2) + (-1)(k) = 6 - k. The dot product is a scalar quantity that reflects the degree to which two vectors point in the same direction. A positive dot product indicates an acute angle between the vectors, a negative dot product an obtuse angle, and a zero dot product orthogonality.

2. Calculate the magnitudes of vectors x and y: The magnitude of a vector is its length. For x, the magnitude ||x|| is ${\sqrt{3^2 + (-1)^2} = \sqrt{10}}$. For y, the magnitude ||y|| is ${\sqrt{2^2 + k^2} = \sqrt{4 + k^2}}$. The magnitude is always a non-negative scalar, representing the vector's length in the given space. Understanding how to calculate the magnitude is essential for many vector-related problems.

3. Use the dot product formula: The dot product formula relates the dot product of two vectors to their magnitudes and the cosine of the angle between them: x · y = ||x|| ||y|| cos θ. We are given that cos θ = ${\frac{\sqrt{5}}{5}}$. This formula bridges the algebraic definition of the dot product with its geometric interpretation, allowing us to connect the angle between vectors with their components and magnitudes.

4. Substitute the calculated values into the formula: Substituting the values we found in steps 1 and 2, and the given cosine value, into the dot product formula, we get: 6 - k = ${\sqrt{10} \cdot \sqrt{4 + k^2} \cdot \frac{\sqrt{5}}{5}}$. This substitution sets up an equation that involves k, which we can then solve to find the possible values of k. It's a critical step in linking the given information to the unknown variable.

5. Solve the equation for k: The equation 6 - k = ${\sqrt{10} \cdot \sqrt{4 + k^2} \cdot \frac{\sqrt{5}}{5}}$ can be simplified and solved for k. This will likely involve squaring both sides of the equation to eliminate the square roots, resulting in a quadratic equation in k. The solutions to this quadratic equation will be the possible values of the constant k. Solving for k is the heart of the problem, requiring careful algebraic manipulation and attention to detail.

Detailed Solution

Let's follow the steps outlined in the methodology to find the values of the constant k.

Step 1: Calculate the dot product of vectors x and y

Given x = 3i - j and y = 2i + kj*, the dot product x · y is calculated as follows:

x · y = (3)(2) + (-1)(k) = 6 - k

This dot product is a scalar value that depends on the constant k. It will play a crucial role in relating the vectors to the cosine of the angle between them. The dot product is a fundamental operation in vector algebra, providing a way to quantify the alignment of two vectors.

Step 2: Calculate the magnitudes of vectors x and y

The magnitude of x, denoted as ||x||, is:

||x|| = ${\sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}}$

The magnitude of y, denoted as ||y||, is:

||y|| = ${\sqrt{2^2 + k^2} = \sqrt{4 + k^2}}$

These magnitudes represent the lengths of the vectors x and y, respectively. They are essential components in the dot product formula, which relates the dot product to the cosine of the angle between the vectors. Magnitudes are always non-negative and provide a measure of the vector's size.

Step 3: Use the dot product formula

The dot product formula states:

x · y = ||x|| ||y|| cos θ

We are given that cos θ = ${\frac{\sqrt{5}}{5}}$. This formula is the bridge between the algebraic definition of the dot product and its geometric interpretation. It allows us to connect the angle between the vectors with their components and magnitudes.

Step 4: Substitute the calculated values into the formula

Substituting the values we calculated in steps 1 and 2, and the given cosine value, into the dot product formula, we get:

6 - k = ${\sqrt{10} \cdot \sqrt{4 + k^2} \cdot \frac{\sqrt{5}}{5}}$

This equation relates the constant k to the dot product, the magnitudes of the vectors, and the cosine of the angle between them. It's a crucial step in setting up the equation that we will solve for k. The equation encapsulates the geometric and algebraic relationships in the problem.

Step 5: Solve the equation for k

To solve the equation 6 - k = ${\sqrt{10} \cdot \sqrt{4 + k^2} \cdot \frac{\sqrt{5}}{5}}$, we first simplify the equation:

6 - k = ${\frac{\sqrt{50(4 + k^2)}}{5}}$

6 - k = ${\frac{\sqrt{200 + 50k^2}}{5}}$

Now, multiply both sides by 5:

5(6 - k) = ${\sqrt{200 + 50k^2}}$

30 - 5k = ${\sqrt{200 + 50k^2}}$

Square both sides to eliminate the square root:

(30 - 5k)^2 = 200 + 50k^2

900 - 300k + 25k^2 = 200 + 50k^2

Rearrange the equation to form a quadratic equation in k:

25k^2 + 300k - 700 = 0

Divide the equation by 25 to simplify:

k^2 + 12k - 28 = 0

Now, we can solve this quadratic equation for k. We can use the quadratic formula or try to factor the equation. Let's try factoring:

(k + 14)(k - 2) = 0

So, the possible values for k are:

k = -14 or k = 2

However, we must check these solutions in the original equation to ensure they are valid, as squaring both sides can introduce extraneous solutions.

Check k = 2:

6 - 2 = ${\sqrt{10} \cdot \sqrt{4 + 2^2} \cdot \frac{\sqrt{5}}{5}}$

4 = ${\sqrt{10} \cdot \sqrt{8} \cdot \frac{\sqrt{5}}{5}}$

4 = ${\frac{\sqrt{400}}{5}}$

4 = ${\frac{20}{5}}$

4 = 4 (This solution is valid)

Check k = -14:

6 - (-14) = ${\sqrt{10} \cdot \sqrt{4 + (-14)^2} \cdot \frac{\sqrt{5}}{5}}$

20 = ${\sqrt{10} \cdot \sqrt{200} \cdot \frac{\sqrt{5}}{5}}$

20 = ${\frac{\sqrt{10000}}{5}}$

20 = ${\frac{100}{5}}$

20 = 20 (This solution is also valid)

Therefore, the values of the constant k are -14 and 2. This thorough check ensures that our solutions are not extraneous and satisfy the original problem conditions. Solving quadratic equations and verifying solutions are crucial skills in mathematical problem-solving.

Conclusion

In this article, we successfully determined the values of the constant k given two vectors x and y and the cosine of the angle between them. By applying the dot product formula and solving the resulting quadratic equation, we found that the possible values for k are -14 and 2. This problem highlights the interplay between vector algebra and trigonometry, demonstrating how these mathematical tools can be used to solve geometric problems in a vector space. The process involved calculating dot products, vector magnitudes, and using the relationship between them to form an equation. We then solved this equation, carefully checking for extraneous solutions to ensure the validity of our answers. Understanding these concepts is crucial for anyone delving into advanced mathematics, physics, or engineering. Vector algebra provides a powerful framework for describing and manipulating quantities with both magnitude and direction, making it an indispensable tool in many scientific and technical fields. The solution not only provides the values of k but also reinforces the importance of careful algebraic manipulation and solution verification in mathematical problem-solving.