Vector Math: Finding The Error In 3u - 5(v + W)

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Hey guys! So, we've got this problem involving vector operations, and someone named Kelcy totally messed it up. The goal is to calculate $3u - 5(v + w)$ with the vectors $u = \langle 9,-2\rangle$, $v = \langle-4,3\rangle$, and $w = \langle 5,1\rangle$. Kelcy got $(32,-26)$, which is, uh, not the right answer. Let's dive into her steps and figure out where she went wrong.

Understanding Vector Operations

Before we even look at Kelcy's work, let's quickly recap what we're dealing with here. We've got vectors, which are basically arrows with both direction and magnitude. In this case, they're represented as coordinate pairs, like $\langle x, y\rangle$. We're asked to perform a few operations: scalar multiplication (multiplying a vector by a number) and vector addition (adding two vectors).

When you multiply a vector $\langle x, y\rangle$ by a scalar $c$, you just multiply each component by that scalar: $c\langle x, y\rangle = \langle cx, cy\rangle$. Simple enough, right?

When you add two vectors, say $\langle x_1, y_1\rangle$ and $\langle x_2, y_2\rangle$, you add their corresponding components: $\langle x_1, y_1\rangle + \langle x_2, y_2\rangle = \langle x_1 + x_2, y_1 + y_2\rangle$. Again, pretty straightforward.

Finally, when you subtract vectors, it's just like addition, but with subtraction: $\langle x_1, y_1\rangle - \langle x_2, y_2\rangle = \langle x_1 - x_2, y_1 - y_2\rangle$. Alternatively, you can think of it as adding the negative of the second vector: $\langle x_1, y_1\rangle + (-1)\langle x_2, y_2\rangle = \nThe expression we need to evaluate is $3u - 5(v + w)$. Following the order of operations (PEMDAS/BODMAS, but for vectors!), we first need to handle the part inside the parentheses: $v + w$. Then, we'll multiply that result by $-5$. After that, we calculate $3u$. Finally, we'll subtract the second result from the first.

Let's break down the specific calculations we'll need:

1. Calculate $v + w$: Add the components of vector $v$ and vector $w$.
2. Calculate $5(v + w)$: Multiply the resulting vector from step 1 by the scalar 5.
3. Calculate $3u$: Multiply vector $u$ by the scalar 3.
4. Calculate $3u - 5(v + w)$: Subtract the vector from step 2 from the vector in step 3.

This systematic approach ensures we don't miss any steps and helps pinpoint errors when they occur. Let's see how Kelcy handled these steps.

Analyzing Kelcy's Steps

Alright, let's put on our detective hats and scrutinize Kelcy's work. She started with the expression $3u - 5(v + w)$ and correctly substituted the given vectors:

$\qquad 3u - 5(v + w) = 3\langle 9,-2\rangle - 5[\langle-4,3\rangle + \langle 5,1\rangle]$

So far, so good! The initial setup looks spot on. Now, let's move to the next line where she starts performing the operations:

$\qquad (27,-6) - 5[(-1,4)]$

Okay, let's pause here. This line shows two things happening:

• Scalar multiplication of $u$: Kelcy calculated $3\langle 9,-2\rangle$ as $\langle 27,-6\rangle$. This is absolutely correct. $3 \times 9 = 27$ and $3 \times (-2) = -6$. Great job, Kelcy!
• Vector addition inside the parentheses: Kelcy calculated $\langle-4,3\rangle + \langle 5,1\rangle$ as $\langle -1,4\rangle$. Let's check this:
  • The x-component: $-4 + 5 = 1$. Uh oh. Kelcy got $-1$. This is the first mistake! She subtracted the x-components instead of adding them.
  • The y-component: $3 + 1 = 4$. This part is correct.

So, the vector inside the brackets should have been $\langle 1,4\rangle$, not $\langle -1,4\rangle$. This single error in the vector addition is going to cascade and mess up the rest of the calculation.

Let's look at the next step Kelcy provided:

$\qquad (27,-6) + \langle 5,-20\rangle$

This line seems to be trying to correct the subtraction sign by turning it into an addition. However, it's also attempting to apply the scalar multiplication of 5 to the incorrect vector $\langle -1,4\rangle$. Let's see what she should have done with the correct vector $\langle 1,4\rangle$ and then what she actually did with her incorrect vector $\langle -1,4\rangle$:

• What she should have done (with the correct vector $\langle 1,4\rangle$): Multiply $\langle 1,4\rangle$ by 5. This would give $5\langle 1,4\rangle = \langle 5 \times 1, 5 \times 4\rangle = \langle 5, 20\rangle$. Then the expression would be $\langle 27,-6\rangle - \langle 5, 20\rangle$.
• What she appears to have done (with her incorrect vector $\langle -1,4\rangle$): She seems to have intended to calculate $-5 \times \langle -1,4\rangle$. If she had done this correctly, it would be $-5\langle -1,4\rangle = \langle (-5) \times (-1), (-5) \times 4\rangle = \langle 5, -20\rangle$. However, the line shows $(27,-6) + \langle 5,-20\rangle$. This means she correctly calculated $5\langle -1,4\rangle$ as $\langle 5,-20\rangle$ (a positive 5 this time? Hmm.) and then added it, effectively treating the operation as $3u + 5(v+w)$ instead of $3u - 5(v+w)$, or she made another mistake in multiplying by -5. Let's assume she was trying to get to the final answer and tried to correct signs. This is where things get really messy.

It looks like Kelcy might have:

1. Made a mistake adding $v+w$.
2. Made a mistake with the scalar multiplication of $-5$.
3. Possibly changed the subtraction to an addition.

Let's re-trace the steps carefully to find the exact point of error and the correct solution.

Correcting Kelcy's Steps and Finding the Right Answer

Okay, let's start fresh and do this step-by-step, paying close attention to every detail. We have $u = \langle 9,-2\rangle$, $v = \langle-4,3\rangle$, and $w = \langle 5,1\rangle$. We need to compute $3u - 5(v + w)$.

Step 1: Calculate $v + w$

This is the part inside the parentheses. We add the corresponding components of $v$ and $w$:

$\qquad v + w = \langle-4,3\rangle + \langle 5,1\rangle$

$\qquad v + w = \langle -4 + 5, 3 + 1 \rangle$

$\qquad v + w = \langle 1, 4 \rangle$

This is where Kelcy made her first error. She calculated this as $\langle -1, 4 \rangle$.

Step 2: Calculate $5(v + w)$

Now, we take the result from Step 1 and multiply it by the scalar 5:

$\qquad 5(v + w) = 5\langle 1, 4 \rangle$

$\qquad 5(v + w) = \langle 5 \times 1, 5 \times 4 \rangle$

$\qquad 5(v + w) = \langle 5, 20 \rangle$

Step 3: Calculate $3u$

Next, we multiply vector $u$ by the scalar 3:

$\qquad 3u = 3\langle 9,-2\rangle$

$\qquad 3u = \langle 3 \times 9, 3 \times (-2) \rangle$

$\qquad 3u = \langle 27, -6 \rangle$

Kelcy got this part right in her first step.

Step 4: Calculate $3u - 5(v + w)$

Finally, we subtract the vector from Step 2 from the vector in Step 3:

$\qquad 3u - 5(v + w) = \langle 27, -6 \rangle - \langle 5, 20 \rangle$

To subtract vectors, we subtract their corresponding components:

$\qquad 3u - 5(v + w) = \langle 27 - 5, -6 - 20 \rangle$

$\qquad 3u - 5(v + w) = \langle 22, -26 \rangle$

So, the correct answer is $\langle 22, -26 \rangle$. Kelcy's answer was $\langle 32, -26 \rangle$. Let's see where her final answer might have come from.

Pinpointing Kelcy's Final Error

Kelcy's final answer was $\langle 32, -26 \rangle$. Our correct answer is $\langle 22, -26 \rangle$. Notice that the y-component, $-26$, is the same! This suggests that her error was isolated to the x-component calculation.

Let's look back at her steps:

$\qquad 3u - 5(v + w)$

$\qquad 3\langle 9,-2\rangle - 5[\langle-4,3\rangle + \langle 5,1\rangle]$

$\qquad (27,-6) - 5[(-1,4)]$

Here's where the x-component calculation went wrong. The first term's x-component is $27$. The term being subtracted involves $v+w$, where the x-component should be $-4+5=1$. Kelcy got $-1$. So, the x-calculation she was trying to do was $27 - ( ext{something derived from } -1)$.

If she correctly calculated $-5 \times (-1)$, she would get $+5$. Then, she would have $27 - (-5)$ if she were subtracting, or $27 + 5$ if she had correctly done $-5 imes (-1)$ and was still subtracting. This is getting confusing because of the sign flips.

Let's assume Kelcy correctly calculated $3u = \langle 27, -6 \rangle$ (which she did) and correctly calculated $v+w = \langle -1, 4 \rangle$ (which she didn't, it should be $\langle 1, 4 \rangle$).

Then she would have $\langle 27, -6 \rangle - 5 \langle -1, 4 \rangle$.

This becomes $\langle 27, -6 \rangle - \langle -5, 20 \rangle$.

Subtracting these: $\langle 27 - (-5), -6 - 20 \rangle = \langle 27 + 5, -26 \rangle = \langle 32, -26 \rangle$.

Aha! This matches Kelcy's final answer. So, Kelcy's mistake was twofold:

1. Incorrectly adding the x-components of $v$ and $w$: She got $-4 + 5 = -1$ instead of $1$.
2. Incorrectly handling the subtraction of the scalar multiple: After getting the incorrect $\langle -1, 4 \rangle$, she correctly calculated $5 \times \langle -1, 4 \rangle$ as $\langle -5, 20 \rangle$, but then she needed to subtract this. The subtraction of the x-component is $27 - (-5)$, which correctly yields $32$. However, the error stems from the initial incorrect sum of $v+w$.

It seems the crucial error was indeed in the very first calculation within the parentheses. The rest of her steps, although complicated by the initial mistake, followed a path that led directly to her incorrect result.

Conclusion

So, guys, the moral of the story is to always double-check your arithmetic, especially with negative signs and vector components. Kelcy made a simple addition error early on, and it snowballed. The correct calculation of $3u - 5(v + w)$ yields $\langle 22, -26 \rangle$. Remember to perform vector operations carefully, step by step, and you'll avoid results like Kelcy's $\langle 32, -26 \rangle$!