Let `veca=(1,1,-1), vecb=(5,-3,-3)` and `vecc=(3,-1,2)`. If `vecr` is collinear with `vecc` and has length `(|veca+vecb|)/(2)`, then `vecr` equals

Options

A. `+-3vecc`
B. `+-3/2vecc`
C. `+-vecc`
D. `+-2/3vecc`

Explanation

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the solution systematically.

### Step 1: Define the vectors
We are given:
- \(\vec{a} = (1, 1, -1)\)
- \(\vec{b} = (5, -3, -3)\)
- \(\vec{c} = (3, -1, 2)\)

### Step 2: Calculate \(|\vec{a} + \vec{b}|\)
First, we need to find the vector sum \(\vec{a} + \vec{b}\):
\[ \vec{a} + \vec{b} = (1 + 5, 1 - 3, -1 - 3) = (6, -2, -4) \]

Next, we calculate the magnitude of \(\vec{a} + \vec{b}\):
\[ |\vec{a} + \vec{b}| = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14} \]

### Step 3: Determine the length of \(\vec{r}\)
We are given that the length of \(\vec{r}\) is:
\[ |\vec{r}| = \frac{|\vec{a} + \vec{b}|}{2} = \frac{2\sqrt{14}}{2} = \sqrt{14} \]

### Step 4: Express \(\vec{r}\) in terms of \(\vec{c}\)
Since \(\vec{r}\) is collinear with \(\vec{c}\), we can express \(\vec{r}\) as:
\[ \vec{r} = \lambda \vec{c} \]
where \(\lambda\) is a scalar.

### Step 5: Calculate the magnitude of \(\vec{c}\)
Now, we need to calculate the magnitude of \(\vec{c}\):
\[ |\vec{c}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]

### Step 6: Relate the magnitudes
From the previous steps, we have:
\[ |\vec{r}| = |\lambda \vec{c}| = |\lambda| |\vec{c}| \]
Substituting the known values:
\[ \sqrt{14} = |\lambda| \cdot \sqrt{14} \]

### Step 7: Solve for \(\lambda\)
Dividing both sides by \(\sqrt{14}\) (assuming \(\sqrt{14} \neq 0\)):
\[ 1 = |\lambda| \]
Thus, \(\lambda = 1\) or \(\lambda = -1\).

### Step 8: Find \(\vec{r}\)
Now substituting back for \(\vec{r}\):
- If \(\lambda = 1\):
\[ \vec{r} = 1 \cdot \vec{c} = (3, -1, 2) \]
- If \(\lambda = -1\):
\[ \vec{r} = -1 \cdot \vec{c} = (-3, 1, -2) \]

### Final Result
Thus, \(\vec{r}\) can be either:
\[ \vec{r} = (3, -1, 2) \quad \text{or} \quad \vec{r} = (-3, 1, -2) \]