If `vec(a)=(2,1,-1), vec(b)=(1,-1,0), vec(c)=(5, -1,1)`, then what is the unit vector parallel to `vec(a)+vec(b)-vec(c)` in the opposite direction ?

Options

A. `(hat(i)+hat(j)-2hat(k))/(3)`
B. `(hat(i)-2hat(j)+2hat(k))/(3)`
C. `(2hat(i)-hat(j)+2hat(k))/(3)`
D. None of the above

Explanation

To solve the problem, we need to find the unit vector parallel to the vector \( \vec{a} + \vec{b} - \vec{c} \) in the opposite direction. Let's break it down step by step.

### Step 1: Define the vectors
We have the following vectors:
- \( \vec{a} = (2, 1, -1) \)
- \( \vec{b} = (1, -1, 0) \)
- \( \vec{c} = (5, -1, 1) \)

### Step 2: Calculate \( \vec{a} + \vec{b} - \vec{c} \)
We need to perform the vector addition and subtraction:
\[ \vec{a} + \vec{b} - \vec{c} = (2, 1, -1) + (1, -1, 0) - (5, -1, 1) \]

Calculating this step by step:
1. First, add \( \vec{a} \) and \( \vec{b} \):
\[ (2 + 1, 1 - 1, -1 + 0) = (3, 0, -1) \]

2. Now subtract \( \vec{c} \):
\[ (3 - 5, 0 - (-1), -1 - 1) = (-2, 1, -2) \]

So, \( \vec{a} + \vec{b} - \vec{c} = (-2, 1, -2) \).

### Step 3: Find the magnitude of the vector
To find the unit vector, we first need the magnitude of \( \vec{v} = (-2, 1, -2) \):
\[ \|\vec{v}\| = \sqrt{(-2)^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 4: Find the unit vector in the same direction
The unit vector \( \hat{v} \) in the direction of \( \vec{v} \) is given by:
\[ \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{(-2, 1, -2)}{3} = \left(-\frac{2}{3}, \frac{1}{3}, -\frac{2}{3}\right) \]

### Step 5: Find the unit vector in the opposite direction
To find the unit vector in the opposite direction, we simply negate the unit vector:
\[ -\hat{v} = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right) \]

### Final Answer
Thus, the unit vector parallel to \( \vec{a} + \vec{b} - \vec{c} \) in the opposite direction is:
\[ \boxed{\left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)} \]