Collinear Vectors: Finding Alpha With Vector Operations

Hey guys! Let's dive into a cool math problem involving vectors. We're going to figure out when two vectors are collinear, which basically means they lie on the same line or are parallel to each other. This is a fundamental concept in linear algebra, and understanding it will give you a solid foundation for more advanced topics. So, buckle up, and let's get started. We will learn how to use vector addition and scalar multiplication to solve for the unknown scalar α (alpha) that makes our vectors collinear. This type of problem often pops up in exams and quizzes, so paying attention is totally worth your while.

Understanding the Problem and the Given Vectors

First off, let's break down the problem. We're given three vectors: $\vec{a}(2; 3)$, $\vec{b}(1; -3)$, and $\vec{c}(-1; 3)$. The notation (x; y) represents the components of the vector in a 2D space. The problem states that two new vectors, $\vec{p}$ and $\vec{q}$, are created using these original vectors and a scalar, $\alpha$ (alpha). $\vec{p}$ is defined as $\vec{a} + \alpha\vec{b}$, and $\vec{q}$ is defined as $\vec{a} + 2\vec{c}$. We need to find the specific value of $\alpha$ that makes $\vec{p}$ and $\vec{q}$ collinear. This means that they must be parallel; the direction of both vectors is the same.

To visualize: Imagine vectors as arrows. If two arrows are collinear, they point in the same direction (or the exact opposite) and lie on the same line. They can have different lengths, but they have to be multiples of each other. This is the key to solving this problem.

Step-by-Step Solution

Alright, let's get down to the nitty-gritty and solve this problem step-by-step. To tackle this, we need to understand a few basic principles: vector addition, scalar multiplication, and the condition for collinearity. Vector addition and scalar multiplication are the main operations we'll need to know. Remember, when we're dealing with vectors, each component is treated separately. If the question gives the vector components, we can easily find the solutions to figure out the value of α.

1. Calculate $\vec{p}$: We know $\vec{p} = \vec{a} + \alpha\vec{b}$. Let's substitute the given values: $\vec{p} = (2; 3) + \alpha(1; -3)$. This simplifies to $\vec{p} = (2 + \alpha; 3 - 3\alpha)$.
2. Calculate $\vec{q}$: Similarly, we know $\vec{q} = \vec{a} + 2\vec{c}$. Plugging in the values, we get $\vec{q} = (2; 3) + 2(-1; 3)$. Simplifying, we have $\vec{q} = (2 - 2; 3 + 6)$, which gives us $\vec{q} = (0; 9)$.
3. Collinearity Condition: For $\vec{p}$ and $\vec{q}$ to be collinear, one must be a scalar multiple of the other. This means there must exist a scalar, let's call it k, such that $\vec{p} = k\vec{q}$.
4. Set up the Equations: Using the components we calculated for $\vec{p}$ and $\vec{q}$, we can write the following equations:
   • $2 + \alpha = k * 0$
   • $3 - 3\alpha = k * 9$
5. Solve for $\alpha$: From the first equation, $2 + \alpha = 0$, we immediately get $\alpha = -2$. This is a crucial step! It is a quick and efficient way to figure out the answer.

Now, let's substitute $\alpha = -2$ into the second equation to double-check: $3 - 3(-2) = 9$. This simplifies to $3 + 6 = 9$, which is true. This confirms that our solution is correct. Therefore, the value of $\alpha$ that makes the vectors $\vec{p}$ and $\vec{q}$ collinear is -2.

Detailed Explanation and Tips

Let's go into more detail about how we solved this problem and some tips that can help you when you encounter similar questions in your math journey. The core concept here is understanding the mathematical definition of collinear vectors. Remember that collinear vectors are parallel and can be expressed as scalar multiples of each other. The scalar can be any real number, which includes positive, negative, and zero values. This is important because it dictates how we set up the equations to solve for the unknown variable, such as $\alpha$ in this problem.

For example, if two vectors are $\vec{u} = (x_1, y_1)$ and $\vec{v} = (x_2, y_2)$, then they are collinear if there exists a scalar k such that $\vec{u} = k\vec{v}$. This translates into $x_1 = kx_2$ and $y_1 = ky_2$. This is the foundation of our solution. We used this property to relate the components of $\vec{p}$ and $\vec{q}$ to find the value of $\alpha$.

When calculating the vector components for $\vec{p}$ and $\vec{q}$, it's crucial to perform the vector addition and scalar multiplication carefully. Make sure you correctly distribute the scalar across the vector components. A small mistake in these calculations can lead to the wrong answer. Double-checking each step is always a good practice, especially during tests or exams.

Another helpful tip is to sketch the vectors, if possible. Visualizing the vectors can help you better understand the problem and confirm your solution. You can draw them on graph paper, or use online vector calculators to make sure the direction is correct. Though this might not always be feasible during a test, it's a great tool for understanding the underlying concepts.

Finally, always ensure that your final answer makes sense in the context of the problem. In this case, our value for $\alpha$ resulted in the vectors $\vec{p}$ and $\vec{q}$ being scalar multiples of each other, confirming their collinearity. If you get a solution that doesn't make logical sense, go back and review your steps to see if there were any calculation errors or misunderstandings.

Choosing the Right Answer

So, looking at the multiple-choice options, which one matches our solution? We found that $\alpha = -2$. Looking back at the problem, we can find that 2 is the correct answer. Congratulations, you've solved it!

Final Thoughts

And there you have it! We've successfully navigated the world of collinear vectors. Hopefully, this explanation has helped you understand the concepts involved and how to solve this type of problem. Remember that practice is key, so try working through similar examples to solidify your understanding. Vector problems might seem complicated at first, but with a bit of practice and by following the logical steps, you can master them. Keep at it, and you'll find that these mathematical concepts are actually pretty interesting. If you have any questions, feel free to ask. Cheers!