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

Options

A. `(hat(i) + hat(j) - 2hat(k))/(3)`
B. `(hat(i) - 2hat(j) + 2hat(k))/(3)`
C. `(2 hat(i) - hat(j) + 2hat(k))/(3)`
D. none of these

Answer

To find the unit vector parallel to \( \vec{a} + \vec{b} - \vec{c} \) in the opposite direction, we will follow these steps:

### Step 1: Calculate \( \vec{a} + \vec{b} - \vec{c} \)

Given:
- \( \vec{a} = (2, 1, -1) \)
- \( \vec{b} = (1, -1, 0) \)
- \( \vec{c} = (5, -1, 1) \)

We can compute \( \vec{a} + \vec{b} - \vec{c} \) as follows:

\[ \vec{a} + \vec{b} = (2 + 1, 1 - 1, -1 + 0) = (3, 0, -1) \]

Now subtract \( \vec{c} \):

\[ \vec{a} + \vec{b} - \vec{c} = (3, 0, -1) - (5, -1, 1) = (3 - 5, 0 - (-1), -1 - 1) = (-2, 1, -2) \]

### Step 2: Find the magnitude of \( \vec{a} + \vec{b} - \vec{c} \)

The magnitude \( |\vec{v}| \) of the vector \( \vec{v} = (-2, 1, -2) \) is calculated as follows:

\[ |\vec{v}| = \sqrt{(-2)^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 3: Find the unit vector in the direction of \( \vec{a} + \vec{b} - \vec{c} \)

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 4: Find the unit vector in the opposite direction

To find the unit vector in the opposite direction, we simply multiply the unit vector by -1:

\[ \hat{v}_{\text{opposite}} = -\hat{v} = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right) \]

### Final Answer

The unit vector parallel to \( \vec{a} + \vec{b} - \vec{c} \) in the opposite direction is:

\[ \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right) \]

---