If the vector `vecb` is collinear with the vector `vec a ( 2sqrt2,-1,4) `and `|vecb| =10`, then

Options

A. `a+-b=0`
B. `a+-2b=0`
C. `2a+-b=0`
D. none of these

Explanation

To solve the problem, we need to find the vector \(\vec{b}\) that is collinear with the vector \(\vec{a} = (2\sqrt{2}, -1, 4)\) and has a magnitude of 10.

### Step-by-step Solution:

1. **Understanding Collinearity**:
Since \(\vec{b}\) is collinear with \(\vec{a}\), we can express \(\vec{b}\) as:
\[ \vec{b} = k \vec{a} \]
where \(k\) is a scalar.

2. **Expressing \(\vec{b}\)**:
Substituting \(\vec{a}\) into the equation:
\[ \vec{b} = k(2\sqrt{2}, -1, 4) = (2\sqrt{2}k, -k, 4k) \]

3. **Finding the Magnitude of \(\vec{b}\)**:
The magnitude of \(\vec{b}\) is given as 10:
\[ |\vec{b}| = \sqrt{(2\sqrt{2}k)^2 + (-k)^2 + (4k)^2} = 10 \]

4. **Calculating the Magnitude**:
Squaring both sides to eliminate the square root:
\[ (2\sqrt{2}k)^2 + (-k)^2 + (4k)^2 = 10^2 \]
Expanding the left side:
\[ 8k^2 + k^2 + 16k^2 = 100 \]
Combining like terms:
\[ 25k^2 = 100 \]

5. **Solving for \(k^2\)**:
Dividing both sides by 25:
\[ k^2 = 4 \]
Taking the square root gives:
\[ k = \pm 2 \]

6. **Finding \(\vec{b}\)**:
Substitute \(k\) back into the expression for \(\vec{b}\):
- For \(k = 2\):
\[ \vec{b} = (2\sqrt{2} \cdot 2, -2, 4 \cdot 2) = (4\sqrt{2}, -2, 8) \]
- For \(k = -2\):
\[ \vec{b} = (2\sqrt{2} \cdot -2, -(-2), 4 \cdot -2) = (-4\sqrt{2}, 2, -8) \]

### Final Result:
Thus, the vectors \(\vec{b}\) can be either:
\[ \vec{b} = (4\sqrt{2}, -2, 8) \quad \text{or} \quad \vec{b} = (-4\sqrt{2}, 2, -8) \]