If `(vec(a) + vec(b)) _|_ vec(b)` and `(vec(a) + 2 vec(b))_|_ vec(a)`, then

Options

A. `|vec(a) | = | vec(b) |`
B. `2 | vec(a)| = |vec(b) |`
C. `|vec(a)| = 2 | vec(b) |`
D. `|vec(a) | = sqrt(2) | vec(b) | `

Answer

To solve the problem, we need to analyze the given conditions step by step.

### Step 1: Understand the conditions
We are given two conditions:
1. \((\vec{a} + \vec{b}) \perp \vec{b}\)
2. \((\vec{a} + 2\vec{b}) \perp \vec{a}\)

### Step 2: Use the property of perpendicular vectors
If two vectors are perpendicular, their dot product is zero. Therefore, we can write:
1. \((\vec{a} + \vec{b}) \cdot \vec{b} = 0\)
2. \((\vec{a} + 2\vec{b}) \cdot \vec{a} = 0\)

### Step 3: Expand the dot products
Now, we will expand both dot products:
1. \((\vec{a} + \vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 0\)
2. \((\vec{a} + 2\vec{b}) \cdot \vec{a} = \vec{a} \cdot \vec{a} + 2\vec{b} \cdot \vec{a} = 0\)

### Step 4: Set up the equations
From the expansions, we can set up the following equations:
1. \(\vec{a} \cdot \vec{b} + |\vec{b}|^2 = 0\) (Equation 1)
2. \(|\vec{a}|^2 + 2\vec{b} \cdot \vec{a} = 0\) (Equation 2)

### Step 5: Solve Equation 1 for \(\vec{a} \cdot \vec{b}\)
From Equation 1, we can express \(\vec{a} \cdot \vec{b}\):
\[ \vec{a} \cdot \vec{b} = -|\vec{b}|^2 \]

### Step 6: Substitute \(\vec{a} \cdot \vec{b}\) into Equation 2
Now substitute \(\vec{a} \cdot \vec{b}\) into Equation 2:
\[ |\vec{a}|^2 + 2(-|\vec{b}|^2) = 0 \]
This simplifies to:
\[ |\vec{a}|^2 - 2|\vec{b}|^2 = 0 \]

### Step 7: Solve for \(|\vec{a}|^2\)
Rearranging gives us:
\[ |\vec{a}|^2 = 2|\vec{b}|^2 \]

### Step 8: Take the square root
Taking the square root of both sides, we find:
\[ |\vec{a}| = \sqrt{2} |\vec{b}| \]

### Conclusion
Thus, we can conclude that:
\[ \vec{a} = \sqrt{2} \vec{b} \]