Math Problem Statement

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

Solution

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

Given:

  • u=3ik\mathbf{u} = 3\mathbf{i} - \mathbf{k}
  • v=2i+3j4k\mathbf{v} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}
  • w=i+5j3k\mathbf{w} = -\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}

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

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

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

Next, add u\mathbf{u} to 2jk2\mathbf{j} - \mathbf{k}. In component form:

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

Now, perform the vector addition:

u+2jk=(3,0,1)+(0,2,1)=(3+0,0+2,11)=(3,2,2)\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 i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} notation)

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

(3,2,2)(3, 2, -2)

c) Compute 2u3w2\mathbf{u} - 3\mathbf{w}

First, find 2u2\mathbf{u} and 3w3\mathbf{w}:

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

3w=3(i+5j3k)=(3i,15j,9k)=(3,15,9)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 3w3\mathbf{w} from 2u2\mathbf{u}:

2u3w=(6,0,2)(3,15,9)=(6(3),015,2(9))=(6+3,15,7)=(9,15,7)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 x\mathbf{x} is a vector such that 13x+u=v+w\frac{1}{3} \mathbf{x} + \mathbf{u} = \mathbf{v} + \mathbf{w}, find x\mathbf{x}

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

= (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.

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Math Problem Analysis

Mathematical Concepts

Vectors
Vector Addition
Scalar Multiplication

Formulas

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Theorems

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Suitable Grade Level

Undergraduate