Let u = 3i − k, let v = 2i + 3j − 4k, and let w = −i + 5j − 3k. a) Compute u + 2j − k. b) Write the answer to part (a) in coordinates (i.e. without using i, j, knotation). c) Compute 2u − 3w. d) If x is a vector such that 1 3 x + u = v + w, find x

Let's tackle each part of the problem step by step.

### Given:
- \(\mathbf{u} = 3\mathbf{i} - \mathbf{k}\)
- \(\mathbf{v} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\)
- \(\mathbf{w} = -\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}\)

### a) Compute \(\mathbf{u} + 2\mathbf{j} - \mathbf{k}\)

First, express \(\mathbf{u}\) in component form:

\[
\mathbf{u} = 3\mathbf{i} + 0\mathbf{j} - \mathbf{k} = (3, 0, -1)
\]

Next, add \(\mathbf{u}\) to \(2\mathbf{j} - \mathbf{k}\). In component form:

\[
2\mathbf{j} - \mathbf{k} = (0, 2, -1)
\]

Now, perform the vector addition:

\[
\mathbf{u} + 2\mathbf{j} - \mathbf{k} = (3, 0, -1) + (0, 2, -1) = (3+0, 0+2, -1-1) = (3, 2, -2)
\]

### b) Write the answer to part (a) in coordinates (i.e., without using \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) notation)

The answer from part (a) is already in coordinate form:

\[
(3, 2, -2)
\]

### c) Compute \(2\mathbf{u} - 3\mathbf{w}\)

First, find \(2\mathbf{u}\) and \(3\mathbf{w}\):

\[
2\mathbf{u} = 2(3\mathbf{i} - \mathbf{k}) = (6\mathbf{i}, 0\mathbf{j}, -2\mathbf{k}) = (6, 0, -2)
\]

\[
3\mathbf{w} = 3(-\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}) = (-3\mathbf{i}, 15\mathbf{j}, -9\mathbf{k}) = (-3, 15, -9)
\]

Now, subtract \(3\mathbf{w}\) from \(2\mathbf{u}\):

\[
2\mathbf{u} - 3\mathbf{w} = (6, 0, -2) - (-3, 15, -9) = (6 - (-3), 0 - 15, -2 - (-9)) = (6 + 3, -15, 7) = (9, -15, 7)
\]

### d) If \(\mathbf{x}\) is a vector such that \(\frac{1}{3} \mathbf{x} + \mathbf{u} = \mathbf{v} + \mathbf{w}\), find \(\mathbf{x}\)

Start by calculating \(\mathbf{v} + \mathbf{w}\):

\[
\mathbf{v} + \mathbf{w} = (2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}) + (-\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}) 
= (2 - 1, 3 + 5, -4 - 3) 
= (1, 8, -7)
\]

Given:

\[
\frac{1}{3} \mathbf{x} + \mathbf{u} = (1, 8, -7)
\]

We know \(\mathbf{u} = (3, 0, -1)\), so:

\[
\frac{1}{3} \mathbf{x} + (3, 0, -1) = (1, 8, -7)
\]

Subtract \(\mathbf{u}\) from both sides:

\[
\frac{1}{3} \mathbf{x} = (1, 8, -7) - (3, 0, -1) = (1 - 3, 8 - 0, -7 + 1) = (-2, 8, -6)
\]

Multiply by 3 to solve for \(\mathbf{x}\):

\[
\mathbf{x} = 3(-2, 8, -6) = (-6, 24, -18)
\]

---

**Summary of Results:**
- a) \(\mathbf{u} + 2\mathbf{j} - \mathbf{k} = (3, 2, -2)\)
- b) \((3, 2, -2)\)
- c) \(2\mathbf{u} - 3\mathbf{w} = (9, -15, 7)\)
- d) \(\mathbf{x} = (-6, 24, -18)\)

Do you have any further questions or need more details?

### Relative Questions:
1. How do you compute the magnitude of a vector like \(\mathbf{x}\) from part (d)?
2. What is the geometric interpretation of adding and subtracting vectors?
3. Can you solve for a vector \(\mathbf{y}\) if given an equation like \(\mathbf{y} + 2\mathbf{z} = \mathbf{a} - \mathbf{b}\)?
4. What happens if you multiply a vector by a negative scalar, such as \(-2\mathbf{u}\)?
5. How would you find a unit vector in the direction of \(\mathbf{u}\)?

### Tip:
Always check your arithmetic carefully when adding or subtracting vectors, especially when working with multiple operations like scalar multiplication and addition.

Explore vector operations involving vectors u, v, w, and solving for vector x in a given vector space problem. Detailed step-by-step solutions provided for each part of the problem, suitable for undergraduate level studies.

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