Solving For K: Vector Magnitude Equation

Hey math enthusiasts! Let's dive into a neat little problem involving vectors and their magnitudes. The question is: If $\|-3 \overrightarrow{A}\|=\|2 k \overrightarrow{ A }\|$, then what's the value of $k$? This kind of problem often pops up in your math journey, so it's super important to understand how to solve it. Don't worry, it's not as scary as it looks. We'll break it down step by step, making sure you grasp every bit of the solution. So, grab your pencils and let's get started. By the end of this, you'll be acing these vector magnitude questions like a pro! This is a classic example of how understanding the properties of vectors can help you solve equations. The core concept here is understanding how scalar multiplication affects the magnitude of a vector. This knowledge is fundamental in linear algebra and has wide applications in fields like physics and computer graphics. Remember that the magnitude of a vector is always a non-negative value; it represents the length or size of the vector.

Let's clarify what's going on here. We have a vector, represented as $\overrightarrow{A}$, and we're dealing with its magnitude, which is the length or size of the vector. The vertical bars around the vector, like $\|\overrightarrow{A}\|$, denote the magnitude. Now, we're given the equation $\|-3 \overrightarrow{A}\|=\|2 k \overrightarrow{ A }\|$. This equation tells us that the magnitude of $-3$ times vector $\overrightarrow{A}$ is equal to the magnitude of $2k$ times the same vector. The key here is to realize how scalar multiplication affects the magnitude of a vector. When you multiply a vector by a scalar (a regular number), you're essentially scaling the vector. If the scalar is negative, you also reverse the direction of the vector. But when we talk about the magnitude, direction doesn't matter; we're only interested in the length. So, if we multiply a vector by a scalar, the magnitude of the resulting vector is the absolute value of the scalar multiplied by the original magnitude. This is where absolute values come into play. Understanding absolute values is crucial to solving the given equation. It's also important to remember the properties of vector magnitudes and how scalar multiplication interacts with them. This ensures that you can confidently solve similar problems. We're going to break down the given equation and simplify it to find the value of $k$. Let's start by recalling what the magnitude of a scalar multiplied by a vector means and applying that to the left side of the equation. We will then simplify the right side of the equation using the same principle. Finally, we'll solve for $k$ to find the value that satisfies the original equation. Let’s make sure we have all the pieces and get them organized to get to the answer.

Understanding Vector Magnitudes and Scalar Multiplication

Alright, before we get to the solution, let's make sure we're all on the same page about vector magnitudes and scalar multiplication. This is super important! Imagine you have a vector $\overrightarrow{A}$. Its magnitude, written as $\|\overrightarrow{A}\|$, is simply its length. Now, if you multiply this vector by a scalar (a number), you're essentially stretching or shrinking the vector. For example, if you multiply $\overrightarrow{A}$ by 2, you get a vector that's twice as long, pointing in the same direction. If you multiply by -2, you get a vector twice as long, but pointing in the opposite direction. But here’s the kicker: when we talk about the magnitude of the resulting vector, we only care about its length, not its direction. The magnitude of $2\overrightarrow{A}$ is $2 \cdot \|\overrightarrow{A}\|$, and the magnitude of $-2\overrightarrow{A}$ is also $2 \cdot \|\overrightarrow{A}\|$. This is because the magnitude is always positive (or zero). So, the key takeaway is: when a vector is multiplied by a scalar, the magnitude of the resulting vector is the absolute value of the scalar times the original magnitude. This is the foundation upon which we’ll solve the problem. Let’s get into a bit more detail. Let’s say we have the vector $\overrightarrow{B}$ and a scalar $c$. The magnitude of $c\overrightarrow{B}$ is given by $\|c\overrightarrow{B}\| = |c| \cdot \|\overrightarrow{B}\|$. The absolute value of the scalar ensures that the magnitude is always non-negative. It means we don't care about the direction; we only care about how much the vector has been stretched or shrunk. Now, think about the equation $\|-3 \overrightarrow{A}\|=\|2 k \overrightarrow{ A }\|$. We can apply this principle to both sides. On the left side, $\|-3 \overrightarrow{A}\|$ becomes $|-3| \cdot \|\overrightarrow{A}\|$, which simplifies to $3 \cdot \|\overrightarrow{A}\|$. On the right side, $\|2 k \overrightarrow{ A }\|$ becomes $|2k| \cdot \|\overrightarrow{A}\|$. So, our equation simplifies to $3 \cdot \|\overrightarrow{A}\| = |2k| \cdot \|\overrightarrow{A}\|$.

This principle is a core concept in vector algebra and is fundamental to solving problems of this kind. Let's make sure it sinks in. If $\|\overrightarrow{V}\|=5$, then $\|3\overrightarrow{V}\| = |3| \cdot 5 = 15$ and $\|-4\overrightarrow{V}\| = |-4| \cdot 5 = 20$. In each case, we apply the absolute value to the scalar to find the resulting magnitude. It doesn't matter if the scalar is positive or negative. The magnitude is always positive. Understanding this concept is critical for solving the original problem. Now, armed with this knowledge, we can move forward with confidence and solve the given equation systematically.

Solving the Equation: Step-by-Step Guide

Okay, guys, let's get down to business and solve this equation. We've already laid the groundwork by understanding the properties of vector magnitudes and scalar multiplication. Now, let’s go through the steps. We'll break it down nice and easy, so you can follow along without any trouble.

First, let's rewrite the original equation: $\|-3 \overrightarrow{A}\|=\|2 k \overrightarrow{ A }\|$. Now, using what we know about the magnitude of a scalar multiplied by a vector, we simplify both sides. The left side, $\|-3 \overrightarrow{A}\|$, becomes $|-3| \cdot \|\overrightarrow{A}\|$, which simplifies to $3 \cdot \|\overrightarrow{A}\|$. The right side, $\|2 k \overrightarrow{ A }\|$, becomes $|2k| \cdot \|\overrightarrow{A}\|$. So, our equation now looks like this: $3 \cdot \|\overrightarrow{A}\| = |2k| \cdot \|\overrightarrow{A}\|$. Notice that $\|\overrightarrow{A}\|$ appears on both sides. If $\|\overrightarrow{A}\|$ is not equal to zero, we can divide both sides by $\|\overrightarrow{A}\|$. Remember, the magnitude of a vector is zero only if the vector itself is a zero vector. So, unless $\overrightarrow{A}$ is the zero vector, we can divide both sides by $\|\overrightarrow{A}\|$ to get $3 = |2k|$. Now we have a simple absolute value equation to solve: $3 = |2k|$. The next step is to remember that the absolute value of a number is its distance from zero. This means that if $|2k| = 3$, then $2k$ could be either $3$ or $-3$. So, we have two possibilities to consider. Let's solve for each one.

Case 1: $2k = 3$. To find $k$, we divide both sides by 2: $k = \frac{3}{2}$.

Case 2: $2k = -3$. Again, to find $k$, we divide both sides by 2: $k = \frac{-3}{2}$.

So, we have two possible solutions for $k$: $k = \frac{3}{2}$ or $k = \frac{-3}{2}$. Let’s ensure we’re on the right track. Remember, the original equation was $\|-3 \overrightarrow{A}\|=\|2 k \overrightarrow{ A }\|$. We know that magnitudes are always positive or zero. Now, when we plug in our values of $k$, we must ensure that the magnitudes are equal. So we can check our answers to make sure we're on the right track. By using this method, you can ensure that you solve the equations correctly. If we plugged in $k = \frac{3}{2}$, the right-hand side becomes $\|2 \cdot \frac{3}{2} \overrightarrow{A}\| = \|3 \overrightarrow{A}\|$, which equals $3 \cdot \|\overrightarrow{A}\|$. This is the same as the left-hand side, since $\|-3 \overrightarrow{A}\| = 3 \cdot \|\overrightarrow{A}\|$. If we plugged in $k = \frac{-3}{2}$, the right-hand side becomes $\|2 \cdot \frac{-3}{2} \overrightarrow{A}\| = \|-3 \overrightarrow{A}\|$, which equals $3 \cdot \|\overrightarrow{A}\|$. This is also the same as the left-hand side. Both values of $k$ satisfy the original equation, meaning both are correct solutions.

Conclusion: The Final Answer

Awesome, you've reached the finish line! So, let's recap. We started with the equation $\|-3 \overrightarrow{A}\|=\|2 k \overrightarrow{ A }\|$ and systematically worked through it. We used the properties of vector magnitudes and scalar multiplication to simplify the equation. We found that the possible values for $k$ are $\frac{3}{2}$ and $\frac{-3}{2}$. These two values of $k$ satisfy the original equation. Thus, the solution to the problem is that k can be either $\frac{3}{2}$ or $\frac{-3}{2}$. Boom! You've successfully solved for $k$. This type of question is fundamental in mathematics, especially in linear algebra. Now, you should be able to solve similar problems with ease. This kind of problem isn't just about finding an answer; it's about understanding the underlying mathematical principles. Keep practicing, and you'll become a vector expert in no time! Also, you've improved your understanding of how scalars affect vector magnitudes and how to solve absolute value equations. So, next time you come across a similar problem, you'll know exactly what to do. Remember, with practice, you'll master these concepts and will be well-prepared for any math challenge that comes your way. Keep up the great work! You've got this! Now, go forth and conquer more math problems. You're well-equipped with the knowledge and skills to handle them. Happy solving! And always remember to double-check your work; it's a good habit to develop. With a solid grasp of vector magnitudes and scalar multiplication, you're well on your way to mastering more complex mathematical concepts. Don't be afraid to practice and ask questions if you get stuck; learning is a journey, and every step counts. Congratulations on solving this problem! Keep up the enthusiasm and continue exploring the fascinating world of mathematics. The journey to mathematical proficiency is exciting, so embrace it and enjoy the process. You are building a strong foundation with each problem you solve. Great job! Keep up the good work! And good luck on your next math adventure. You're doing great! Keep practicing and expanding your knowledge to excel in all your math endeavors. Well done!