Solving Vector Equations: A Step-by-Step Guide

Hey guys! Let's dive into some vector equations. We're going to break down how to find a vector $\vec{x}$ given points and other vectors. This stuff is super useful in a bunch of areas, from physics to computer graphics. We'll go through the problem step-by-step, so it's easy to follow along. Ready?

Part A: Unraveling the First Equation

So, the first part of our problem gives us the equation: $3\vec{x} + 2\vec{v} = \vec{x} + (\vec{AB} \cdot \vec{u})\vec{v}$. Our goal is to find $\vec{x}$. The given points are $A(4, 0, -1)$, $B(2, -2, 1)$, and $C(1, 3, 2)$, and the vectors are $\vec{u} = (2, 1, 1)$ and $\vec{v} = (-1, -2, 3)$. Let's break this down piece by piece. It's like a puzzle, and we're putting the pieces together!

First, we need to find the vector $\vec{AB}$. Remember how to do that? We subtract the coordinates of point A from the coordinates of point B. So, $\vec{AB} = B - A = (2 - 4, -2 - 0, 1 - (-1)) = (-2, -2, 2)$. Cool, huh?

Next, we need to calculate the dot product of $\vec{AB}$ and $\vec{u}$, which is written as $\vec{AB} \cdot \vec{u}$. The dot product is found by multiplying the corresponding components of the vectors and then adding the results. So, $\vec{AB} \cdot \vec{u} = (-2 \cdot 2) + (-2 \cdot 1) + (2 \cdot 1) = -4 - 2 + 2 = -4$. Easy peasy!

Now we can plug this value back into our original equation: $3\vec{x} + 2\vec{v} = \vec{x} + (-4)\vec{v}$. We've made some real progress! Let's rearrange this equation to isolate $\vec{x}$ terms. We'll move the $\vec{x}$ term from the right side to the left side, and the $2\vec{v}$ term from the left side to the right side. This gives us: $3\vec{x} - \vec{x} = -4\vec{v} - 2\vec{v}$. Simplify it, and we get: $2\vec{x} = -6\vec{v}$.

Almost there! Now, to solve for $\vec{x}$, we just need to divide both sides of the equation by 2. Therefore, $\vec{x} = -3\vec{v}$. Remember that $\vec{v} = (-1, -2, 3)$. So, to find the final vector $\vec{x}$, we multiply $\vec{v}$ by -3: $\vec{x} = -3(-1, -2, 3) = (3, 6, -9)$. And that's the answer for the first part! High five! We've successfully solved for $\vec{x}$ in the first equation! It's like solving a riddle; each step brings us closer to the solution. We used vector subtraction, dot products, and some basic algebra. Doesn't that feel awesome?

Summary of Part A:

1. Calculate $\vec{AB}$ from points A and B.
2. Find the dot product of $\vec{AB}$ and $\vec{u}$.
3. Substitute the dot product into the original equation.
4. Rearrange and simplify the equation to isolate $\vec{x}$.
5. Solve for $\vec{x}$ by multiplying by a scalar.

Part B: Tackling the Second Equation

Alright, let's move on to the second equation! We're going to find the value of $\vec{x}$ in the equation: $(\vec{BC} \cdot \vec{x})\vec{u} = \vec{BC} \times \vec{u}$. This one involves the cross product, which is a bit different from the dot product. Let's take it one step at a time. Remember our points: $B(2, -2, 1)$ and $C(1, 3, 2)$, and $\vec{u} = (2, 1, 1)$.

First, let's find $\vec{BC}$. Just like we did before, we subtract the coordinates of point B from point C. So, $\vec{BC} = C - B = (1 - 2, 3 - (-2), 2 - 1) = (-1, 5, 1)$.

Next up, we need to calculate the cross product of $\vec{BC}$ and $\vec{u}$, written as $\vec{BC} \times \vec{u}$. Remember how to do this? You can use the determinant method. It goes like this: $\vec{BC} \times \vec{u} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 5 & 1 \\ 2 & 1 & 1 \end{vmatrix}$.

To solve this determinant, we get: $\hat{i}(5 \cdot 1 - 1 \cdot 1) - \hat{j}(-1 \cdot 1 - 1 \cdot 2) + \hat{k}(-1 \cdot 1 - 5 \cdot 2) = 4\hat{i} + 3\hat{j} - 11\hat{k}$. So, $\vec{BC} \times \vec{u} = (4, 3, -11)$.

Now our equation is: $(\vec{BC} \cdot \vec{x})\vec{u} = (4, 3, -11)$. Notice that the left side of the equation is a scalar multiple of $\vec{u}$. This means that the resulting vector must be parallel to $\vec{u}$. However, the vector on the right side, $(4, 3, -11)$, is not parallel to $\vec{u} = (2, 1, 1)$. The components of $(4, 3, -11)$ do not have the same ratios as those of $(2, 1, 1)$.

This means that for this equation to hold true, the cross product $\vec{BC} \times \vec{u}$ must equal the zero vector $(0, 0, 0)$. However, as we've already calculated, it doesn't. Therefore, there is no solution for $\vec{x}$ that satisfies this equation. Whoa, mind blown! Sometimes, even in math, there are no solutions. It's important to recognize when this happens and understand why. We learned about vector subtraction, the cross product, and when a solution might not exist. It's like a treasure hunt where the treasure isn't there, and now you know how to read the map!

Summary of Part B:

1. Calculate $\vec{BC}$ from points B and C.
2. Find the cross product of $\vec{BC}$ and $\vec{u}$.
3. Analyze the resulting equation.
4. Determine that there is no solution because the vectors are not parallel.

Final Thoughts

And that's a wrap! We've gone through two vector equations, learning how to solve for $\vec{x}$ in each. In the first equation, we successfully found a solution using dot products and algebraic manipulation. However, the second equation revealed that, in some cases, there may be no solution at all. This is a super important concept to understand. These examples illustrate how to use vectors and their properties to solve mathematical problems. Keep practicing, and you'll become a vector master in no time!

Remember, understanding the basics – vector addition, subtraction, dot products, and cross products – is key. The more you practice, the more comfortable you'll become with these concepts. Don't be afraid to revisit the examples and work through them again. Breaking down problems into smaller, manageable steps, will make the process a whole lot easier. Keep up the great work, and happy vectoring!