Math Problem Statement

Let u = (1, 2, 3),

v = (2, 2, −1), and w = (4, −3, −1). Find z, where 2u + v − w + 3z = 0.

Solution

We are given the equation:

2u+vw+3z=02\mathbf{u} + \mathbf{v} - \mathbf{w} + 3\mathbf{z} = 0

where u=(1,2,3)\mathbf{u} = (1, 2, 3), v=(2,2,1)\mathbf{v} = (2, 2, -1), and w=(4,3,1)\mathbf{w} = (4, -3, -1). We need to find z\mathbf{z}.

Step 1: Expand the terms

First, let's compute each vector component separately.

  • 2u=2(1,2,3)=(2,4,6)2\mathbf{u} = 2(1, 2, 3) = (2, 4, 6)
  • v=(2,2,1)\mathbf{v} = (2, 2, -1)
  • w=(4,3,1)\mathbf{w} = (4, -3, -1)

Now substitute these values into the equation:

(2,4,6)+(2,2,1)(4,3,1)+3z=0(2, 4, 6) + (2, 2, -1) - (4, -3, -1) + 3\mathbf{z} = 0

Step 2: Simplify the left-hand side

Start by adding the vectors 2u2\mathbf{u}, v\mathbf{v}, and w-\mathbf{w}:

(2,4,6)+(2,2,1)=(4,6,5)(2, 4, 6) + (2, 2, -1) = (4, 6, 5) (4,6,5)(4,3,1)=(44,6(3),5(1))=(0,9,6)(4, 6, 5) - (4, -3, -1) = (4-4, 6-(-3), 5-(-1)) = (0, 9, 6)

So the equation becomes:

(0,9,6)+3z=0(0, 9, 6) + 3\mathbf{z} = 0

Step 3: Solve for z\mathbf{z}

Now, we want to isolate 3z3\mathbf{z}. Subtract (0,9,6)(0, 9, 6) from both sides:

3z=(0,9,6)3\mathbf{z} = (0, -9, -6)

Finally, divide both sides by 3 to find z\mathbf{z}:

z=(03,93,63)=(0,3,2)\mathbf{z} = \left( \frac{0}{3}, \frac{-9}{3}, \frac{-6}{3} \right) = (0, -3, -2)

Final Answer:

z=(0,3,2)\mathbf{z} = (0, -3, -2)

Would you like more details or have any questions about this solution? Here's some follow-up:

  1. How to compute the magnitude of a vector?
  2. What is the dot product of u\mathbf{u} and v\mathbf{v}?
  3. How to check if two vectors are perpendicular?
  4. What is the cross product of v\mathbf{v} and w\mathbf{w}?
  5. How to find the angle between two vectors?

Tip: When working with vector equations, break down each component separately, and remember to handle scalar multiplications and vector additions/subtractions one component at a time for accuracy.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Linear Equations
Vector Addition
Scalar Multiplication

Formulas

Vector addition: (a1, a2, a3) + (b1, b2, b3) = (a1 + b1, a2 + b2, a3 + b3)
Scalar multiplication: c(a1, a2, a3) = (ca1, ca2, ca3)
Solving for unknown vector: Ax + B = 0 ⟹ x = -B/A

Theorems

Properties of Vector Addition and Scalar Multiplication

Suitable Grade Level

Grades 10-12