A vector b collinear with `a=2sqrt(2)i-j+4k` of length 10 is given by

Options

A. `3(2sqrt(2)i-j+4k)`
B. `2(2sqrt(2)i+j-4k)`
C. `2(2sqrt(2)i+j+4k)`
D. `2(2sqrt(2)i-j+4k)`

Answer

To find the vector \( \mathbf{b} \) that is collinear with the vector \( \mathbf{a} = 2\sqrt{2} \mathbf{i} - \mathbf{j} + 4\mathbf{k} \) and has a length of 10, we can follow these steps:

### Step 1: Express \( \mathbf{b} \) in terms of \( \mathbf{a} \)
Since \( \mathbf{b} \) is collinear with \( \mathbf{a} \), we can express \( \mathbf{b} \) as:
\[ \mathbf{b} = \lambda \mathbf{a} \]
where \( \lambda \) is a scalar.

### Step 2: Substitute \( \mathbf{a} \) into the equation
Substituting \( \mathbf{a} \) into the equation gives:
\[ \mathbf{b} = \lambda (2\sqrt{2} \mathbf{i} - \mathbf{j} + 4\mathbf{k}) \]

### Step 3: Calculate the magnitude of \( \mathbf{a} \)
To find the magnitude of \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{(2\sqrt{2})^2 + (-1)^2 + 4^2} \]
Calculating each term:
- \( (2\sqrt{2})^2 = 8 \)
- \( (-1)^2 = 1 \)
- \( 4^2 = 16 \)

Adding these:
\[ |\mathbf{a}| = \sqrt{8 + 1 + 16} = \sqrt{25} = 5 \]

### Step 4: Set the magnitude of \( \mathbf{b} \)
We know that the magnitude of \( \mathbf{b} \) is given as 10:
\[ |\mathbf{b}| = 10 \]
Substituting for \( \mathbf{b} \):
\[ |\mathbf{b}| = |\lambda| |\mathbf{a}| \]
Thus,
\[ 10 = |\lambda| \cdot 5 \]

### Step 5: Solve for \( \lambda \)
Dividing both sides by 5:
\[ |\lambda| = \frac{10}{5} = 2 \]
Therefore, \( \lambda \) can be either \( 2 \) or \( -2 \).

### Step 6: Find \( \mathbf{b} \)
Using \( \lambda = 2 \):
\[ \mathbf{b} = 2(2\sqrt{2} \mathbf{i} - \mathbf{j} + 4\mathbf{k}) = 4\sqrt{2} \mathbf{i} - 2\mathbf{j} + 8\mathbf{k} \]

Alternatively, using \( \lambda = -2 \):
\[ \mathbf{b} = -2(2\sqrt{2} \mathbf{i} - \mathbf{j} + 4\mathbf{k}) = -4\sqrt{2} \mathbf{i} + 2\mathbf{j} - 8\mathbf{k} \]

### Final Answer
Thus, the vector \( \mathbf{b} \) can be:
\[ \mathbf{b} = 4\sqrt{2} \mathbf{i} - 2\mathbf{j} + 8\mathbf{k} \quad \text{or} \quad \mathbf{b} = -4\sqrt{2} \mathbf{i} + 2\mathbf{j} - 8\mathbf{k} \]