If` vece_(1) = ( 1,1,1) and vece_(2) = ( 1,1,-1) and veca and vecb` are two vectors that ` vece_(1) = 2veca +vecb and vece_(2) veca + 2vecb` then angle between `veca and vecb` is

Options

A. `cos^(-1)(7/9)`
B. `cos^(-1)(-7/11)`
C. `cos^(-1)(-7/9)`
D. `cos^(-1)(6sqrt(2))/11`

Answer

To find the angle between the vectors \( \vec{a} \) and \( \vec{b} \) given the equations involving vectors \( \vec{e_1} \) and \( \vec{e_2} \), we can follow these steps:

### Step 1: Write down the given vectors
We have:
\[ \vec{e_1} = (1, 1, 1) \quad \text{and} \quad \vec{e_2} = (1, 1, -1) \]

### Step 2: Set up the equations based on the problem statement
From the problem, we know:
\[ \vec{e_1} = 2\vec{a} + \vec{b} \quad \text{(1)} \]
\[ \vec{e_2} = \vec{a} + 2\vec{b} \quad \text{(2)} \]

### Step 3: Substitute the values of \( \vec{e_1} \) and \( \vec{e_2} \)
Substituting the values of \( \vec{e_1} \) and \( \vec{e_2} \) into equations (1) and (2):
From equation (1):
\[ (1, 1, 1) = 2\vec{a} + \vec{b} \]
From equation (2):
\[ (1, 1, -1) = \vec{a} + 2\vec{b} \]

### Step 4: Express \( \vec{b} \) in terms of \( \vec{a} \)
From equation (1), we can express \( \vec{b} \):
\[ \vec{b} = (1, 1, 1) - 2\vec{a} \quad \text{(3)} \]

### Step 5: Substitute \( \vec{b} \) into equation (2)
Now, substitute equation (3) into equation (2):
\[ (1, 1, -1) = \vec{a} + 2((1, 1, 1) - 2\vec{a}) \]
Expanding this gives:
\[ (1, 1, -1) = \vec{a} + 2(1, 1, 1) - 4\vec{a} \]
\[ (1, 1, -1) = (2, 2, 2) - 3\vec{a} \]

### Step 6: Rearranging to find \( \vec{a} \)
Rearranging gives:
\[ 3\vec{a} = (2, 2, 2) - (1, 1, -1) \]
\[ 3\vec{a} = (1, 1, 3) \]
\[ \vec{a} = \left(\frac{1}{3}, \frac{1}{3}, 1\right) \]

### Step 7: Substitute \( \vec{a} \) back to find \( \vec{b} \)
Now substitute \( \vec{a} \) back into equation (3) to find \( \vec{b} \):
\[ \vec{b} = (1, 1, 1) - 2\left(\frac{1}{3}, \frac{1}{3}, 1\right) \]
\[ \vec{b} = (1, 1, 1) - \left(\frac{2}{3}, \frac{2}{3}, 2\right) \]
\[ \vec{b} = \left(1 - \frac{2}{3}, 1 - \frac{2}{3}, 1 - 2\right) = \left(\frac{1}{3}, \frac{1}{3}, -1\right) \]

### Step 8: Find the angle between \( \vec{a} \) and \( \vec{b} \)
Now we can find the angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \) using the dot product formula:
\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]

### Step 9: Calculate the dot product \( \vec{a} \cdot \vec{b} \)
Calculating the dot product:
\[ \vec{a} \cdot \vec{b} = \left(\frac{1}{3}, \frac{1}{3}, 1\right) \cdot \left(\frac{1}{3}, \frac{1}{3}, -1\right) = \frac{1}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{3} + 1 \cdot (-1) \]
\[ = \frac{1}{9} + \frac{1}{9} - 1 = \frac{2}{9} - 1 = \frac{2}{9} - \frac{9}{9} = -\frac{7}{9} \]

### Step 10: Calculate the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \)
Calculating the magnitudes:
\[ |\vec{a}| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + 1^2} = \sqrt{\frac{1}{9} + \frac{1}{9} + 1} = \sqrt{\frac{2}{9} + 1} = \sqrt{\frac{11}{9}} = \frac{\sqrt{11}}{3} \]
\[ |\vec{b}| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + (-1)^2} = \sqrt{\frac{1}{9} + \frac{1}{9} + 1} = \sqrt{\frac{2}{9} + 1} = \sqrt{\frac{11}{9}} = \frac{\sqrt{11}}{3} \]

### Step 11: Substitute into the cosine formula
Substituting into the cosine formula:
\[ \cos \theta = \frac{-\frac{7}{9}}{\left(\frac{\sqrt{11}}{3}\right) \left(\frac{\sqrt{11}}{3}\right)} = \frac{-\frac{7}{9}}{\frac{11}{9}} = -\frac{7}{11} \]

### Step 12: Find the angle \( \theta \)
Finally, we find \( \theta \):
\[ \theta = \cos^{-1}\left(-\frac{7}{11}\right) \]

### Final Answer
The angle between vectors \( \vec{a} \) and \( \vec{b} \) is:
\[ \theta = \cos^{-1}\left(-\frac{7}{11}\right) \]