A vecotr `vec(b)` is collinear with the vector `vec(a)= (2, 1, -1)` and satisfies the condition `vec(a). vec(b)= 3`. What is `vec(b)` equal to ?

Options

A. `(1,1//2, -1//2)`
B. `(2//3.1//3, -1//3)`
C. `(1//2, 1//4, -1//4)`
D. `(1, 1, 0)`

Answer

To solve the problem, we need to find the vector \(\vec{b}\) that is collinear with the vector \(\vec{a} = (2, 1, -1)\) and satisfies the condition \(\vec{a} \cdot \vec{b} = 3\).

### Step 1: Understand Collinearity
Since \(\vec{b}\) is collinear with \(\vec{a}\), we can express \(\vec{b}\) as a scalar multiple of \(\vec{a}\). Therefore, we can write:
\[ \vec{b} = k \vec{a} = k(2, 1, -1) \]
where \(k\) is a scalar.

### Step 2: Substitute \(\vec{b}\) in the Dot Product Condition
Now, we substitute \(\vec{b}\) into the dot product condition:
\[ \vec{a} \cdot \vec{b} = \vec{a} \cdot (k(2, 1, -1)) = 3 \]
Calculating the dot product:
\[ \vec{a} \cdot \vec{b} = (2, 1, -1) \cdot (2k, k, -k) = 2(2k) + 1(k) + (-1)(-k) \]
This simplifies to:
\[ 4k + k + k = 6k \]

### Step 3: Set Up the Equation
Now we set up the equation based on the condition given:
\[ 6k = 3 \]

### Step 4: Solve for \(k\)
To find \(k\), we divide both sides by 6:
\[ k = \frac{3}{6} = \frac{1}{2} \]

### Step 5: Find \(\vec{b}\)
Now that we have \(k\), we can find \(\vec{b}\):
\[ \vec{b} = k \vec{a} = \frac{1}{2}(2, 1, -1) = (1, \frac{1}{2}, -\frac{1}{2}) \]

Thus, the vector \(\vec{b}\) is:
\[ \vec{b} = \left(1, \frac{1}{2}, -\frac{1}{2}\right) \]

### Final Answer
\(\vec{b} = (1, \frac{1}{2}, -\frac{1}{2})\)

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