Math Problem Statement

Solution

The problem given involves vectors a\mathbf{a} and b\mathbf{b} with the following properties:

  1. a=1|\mathbf{a}| = 1
  2. a+2b=2|\mathbf{a} + 2\mathbf{b}| = 2
  3. (b2a)b(\mathbf{b} - 2\mathbf{a}) \perp \mathbf{b}

We need to find the magnitude b|\mathbf{b}|.

Let's solve this step by step.

Step 1: Use the Magnitude Condition

From the first condition: a=1|\mathbf{a}| = 1

Step 2: Use the Second Condition

a+2b=2|\mathbf{a} + 2\mathbf{b}| = 2 Squaring both sides: (a+2b)(a+2b)=4(\mathbf{a} + 2\mathbf{b}) \cdot (\mathbf{a} + 2\mathbf{b}) = 4 aa+4ab+4bb=4\mathbf{a} \cdot \mathbf{a} + 4 \mathbf{a} \cdot \mathbf{b} + 4 \mathbf{b} \cdot \mathbf{b} = 4 Since a=1|\mathbf{a}| = 1, we have aa=1\mathbf{a} \cdot \mathbf{a} = 1: 1+4(ab)+4b2=41 + 4 (\mathbf{a} \cdot \mathbf{b}) + 4 |\mathbf{b}|^2 = 4 Rearranging: 4b2+4(ab)=34 |\mathbf{b}|^2 + 4 (\mathbf{a} \cdot \mathbf{b}) = 3 b2+(ab)=34|\mathbf{b}|^2 + (\mathbf{a} \cdot \mathbf{b}) = \frac{3}{4}

Step 3: Use the Perpendicular Condition

Since (b2a)(\mathbf{b} - 2\mathbf{a}) is perpendicular to b\mathbf{b}: (b2a)b=0(\mathbf{b} - 2\mathbf{a}) \cdot \mathbf{b} = 0 bb2ab=0\mathbf{b} \cdot \mathbf{b} - 2 \mathbf{a} \cdot \mathbf{b} = 0 b22(ab)=0|\mathbf{b}|^2 - 2 (\mathbf{a} \cdot \mathbf{b}) = 0 b2=2(ab)|\mathbf{b}|^2 = 2 (\mathbf{a} \cdot \mathbf{b})

Step 4: Solve the System of Equations

We have the following two equations:

  1. b2+(ab)=34|\mathbf{b}|^2 + (\mathbf{a} \cdot \mathbf{b}) = \frac{3}{4}
  2. b2=2(ab)|\mathbf{b}|^2 = 2 (\mathbf{a} \cdot \mathbf{b})

Substitute 2(ab)2 (\mathbf{a} \cdot \mathbf{b}) for b2|\mathbf{b}|^2 in the first equation: 2(ab)+(ab)=342 (\mathbf{a} \cdot \mathbf{b}) + (\mathbf{a} \cdot \mathbf{b}) = \frac{3}{4} 3(ab)=343 (\mathbf{a} \cdot \mathbf{b}) = \frac{3}{4} ab=14\mathbf{a} \cdot \mathbf{b} = \frac{1}{4}

Now, substitute ab=14\mathbf{a} \cdot \mathbf{b} = \frac{1}{4} into the second equation: b2=2(14)|\mathbf{b}|^2 = 2 \left( \frac{1}{4} \right) b2=12|\mathbf{b}|^2 = \frac{1}{2} b=22|\mathbf{b}| = \frac{\sqrt{2}}{2}

Therefore, the correct answer is: b=22|\mathbf{b}| = \frac{\sqrt{2}}{2}

So, the correct option is B.

Do you want more details or have any questions?

Here are 5 more questions you might want to ask:

  1. How do you determine if two vectors are perpendicular?
  2. Can you explain the dot product in more detail?
  3. What are some common vector operations and their properties?
  4. How do you find the magnitude of a vector given its components?
  5. What are some real-world applications of vector mathematics?

Tip: Always break down complex vector problems into smaller parts and use fundamental vector properties to simplify the solution process.

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Math Problem Analysis

Mathematical Concepts

Vectors
Magnitude of Vectors
Dot Product
Vector Perpendicularity

Formulas

Magnitude of a vector formula
Dot product formula

Theorems

Properties of vector magnitudes
Vector perpendicularity theorem

Suitable Grade Level

Advanced High School