A vector b, which is collinear with vector `a=2i+j-k` and satisfies `a*b=2` is given by

Options

A. `1/2(2i+j-k)`
B. `1/3(2i+j-k)`
C. `1/4(2i+j-k)`
D. `1/2(-2i-j+k)`

Answer

To solve the problem, we need to find the vector \( \mathbf{b} \) that is collinear with the vector \( \mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k} \) and satisfies the condition \( \mathbf{a} \cdot \mathbf{b} = 2 \).

### 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 \( \mathbf{b} \) into the dot product equation
We know that:
\[ \mathbf{a} \cdot \mathbf{b} = 2 \]
Substituting \( \mathbf{b} \) gives:
\[ \mathbf{a} \cdot (\lambda \mathbf{a}) = 2 \]
This simplifies to:
\[ \lambda (\mathbf{a} \cdot \mathbf{a}) = 2 \]

### Step 3: Calculate \( \mathbf{a} \cdot \mathbf{a} \)
Now, we need to compute \( \mathbf{a} \cdot \mathbf{a} \):
\[ \mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k} \]
Calculating the dot product:
\[ \mathbf{a} \cdot \mathbf{a} = (2)^2 + (1)^2 + (-1)^2 = 4 + 1 + 1 = 6 \]

### Step 4: Substitute back to find \( \lambda \)
Now we substitute back into the equation:
\[ \lambda \cdot 6 = 2 \]
Solving for \( \lambda \):
\[ \lambda = \frac{2}{6} = \frac{1}{3} \]

### Step 5: Find \( \mathbf{b} \)
Now we can find \( \mathbf{b} \):
\[ \mathbf{b} = \lambda \mathbf{a} = \frac{1}{3} (2\mathbf{i} + \mathbf{j} - \mathbf{k}) \]
Distributing \( \frac{1}{3} \):
\[ \mathbf{b} = \frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{1}{3} \mathbf{k} \]

### Final Answer
Thus, the vector \( \mathbf{b} \) is:
\[ \mathbf{b} = \frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{1}{3} \mathbf{k} \]

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