If a and b are two vectors such that `2a+b=e_(1)" and "a+2b=e_(2)`, where `e_(1)=(1, 1, 1)" and "e_(2)=(1,1,-1)`, then the angle between a and b is

Options

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

Answer

To solve the problem, we need to find the angle between the two vectors \( \mathbf{a} \) and \( \mathbf{b} \) given the equations:

1. \( 2\mathbf{a} + \mathbf{b} = \mathbf{e}_1 \)
2. \( \mathbf{a} + 2\mathbf{b} = \mathbf{e}_2 \)

where \( \mathbf{e}_1 = (1, 1, 1) \) and \( \mathbf{e}_2 = (1, 1, -1) \).

### Step 1: Express the vectors in terms of \( \mathbf{a} \) and \( \mathbf{b} \)

From the first equation, we can express \( \mathbf{b} \) in terms of \( \mathbf{a} \):

\[ \mathbf{b} = \mathbf{e}_1 - 2\mathbf{a} \]

### Step 2: Substitute \( \mathbf{b} \) into the second equation

Now, substitute \( \mathbf{b} \) into the second equation:

\[ \mathbf{a} + 2(\mathbf{e}_1 - 2\mathbf{a}) = \mathbf{e}_2 \]

Expanding this gives:

\[ \mathbf{a} + 2\mathbf{e}_1 - 4\mathbf{a} = \mathbf{e}_2 \]

Combining like terms results in:

\[ -3\mathbf{a} + 2\mathbf{e}_1 = \mathbf{e}_2 \]

### Step 3: Rearranging to find \( \mathbf{a} \)

Rearranging gives:

\[ 3\mathbf{a} = 2\mathbf{e}_1 - \mathbf{e}_2 \]

Substituting the values of \( \mathbf{e}_1 \) and \( \mathbf{e}_2 \):

\[ 3\mathbf{a} = 2(1, 1, 1) - (1, 1, -1) \]

Calculating this yields:

\[ 3\mathbf{a} = (2, 2, 2) - (1, 1, -1) = (1, 1, 3) \]

Thus,

\[ \mathbf{a} = \left(\frac{1}{3}, \frac{1}{3}, 1\right) \]

### Step 4: Substitute \( \mathbf{a} \) back to find \( \mathbf{b} \)

Now substitute \( \mathbf{a} \) back into the equation for \( \mathbf{b} \):

\[ \mathbf{b} = \mathbf{e}_1 - 2\mathbf{a} = (1, 1, 1) - 2\left(\frac{1}{3}, \frac{1}{3}, 1\right) \]

Calculating this gives:

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

### Step 5: Find the angle between \( \mathbf{a} \) and \( \mathbf{b} \)

To find the angle \( \theta \) between \( \mathbf{a} \) and \( \mathbf{b} \), we use the dot product formula:

\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]

Calculating \( \mathbf{a} \cdot \mathbf{b} \):

\[ \mathbf{a} \cdot \mathbf{b} = \left(\frac{1}{3}, \frac{1}{3}, 1\right) \cdot \left(\frac{1}{3}, \frac{1}{3}, -1\right) = \frac{1}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{3} + 1 \cdot (-1) = \frac{1}{9} + \frac{1}{9} - 1 = \frac{2}{9} - 1 = -\frac{7}{9} \]

Calculating the magnitudes:

\[ |\mathbf{a}| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + 1^2} = \sqrt{\frac{1}{9} + \frac{1}{9} + 1} = \sqrt{\frac{2}{9} + 1} = \sqrt{\frac{11}{9}} = \frac{\sqrt{11}}{3} \]

\[ |\mathbf{b}| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + (-1)^2} = \sqrt{\frac{1}{9} + \frac{1}{9} + 1} = \sqrt{\frac{2}{9} + 1} = \sqrt{\frac{11}{9}} = \frac{\sqrt{11}}{3} \]

### Step 6: Substitute into the angle formula

Substituting into the angle formula gives:

\[ -\frac{7}{9} = \left(\frac{\sqrt{11}}{3}\right) \left(\frac{\sqrt{11}}{3}\right) \cos \theta \]

This simplifies to:

\[ -\frac{7}{9} = \frac{11}{9} \cos \theta \]

Thus,

\[ \cos \theta = -\frac{7}{11} \]

### Step 7: Find \( \theta \)

Finally, we find \( \theta \):

\[ \theta = \cos^{-1}\left(-\frac{7}{11}\right) \]