Find angle `theta`between the vectors ` veca= hat i+ hat j- hat k`and ` vecb= hat i- hat j+ hat k`.

Options

A. `cos^(-1)(1/3)`
B. `cos^(-1)(-1/2)`
C. `cos^(-1)(-1/3)`
D. `cos^(-1)(1/2)`

Explanation

To find the angle \( \theta \) between the vectors \( \vec{a} = \hat{i} + \hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \), we will use the formula for the dot product of two vectors:

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

### Step 1: Calculate the dot product \( \vec{a} \cdot \vec{b} \)

The dot product is calculated as follows:

\[ \vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) \]

Calculating this gives:

\[ \vec{a} \cdot \vec{b} = 1 - 1 - 1 = -1 \]

### Step 2: Calculate the magnitudes of \( \vec{a} \) and \( \vec{b} \)

The magnitude of vector \( \vec{a} \) is:

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

The magnitude of vector \( \vec{b} \) is:

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

### Step 3: Substitute into the dot product formula

Now we substitute the values into the dot product formula:

\[ -1 = |\vec{a}| |\vec{b}| \cos \theta \]

Substituting the magnitudes:

\[ -1 = (\sqrt{3})(\sqrt{3}) \cos \theta \]

This simplifies to:

\[ -1 = 3 \cos \theta \]

### Step 4: Solve for \( \cos \theta \)

Now, we can solve for \( \cos \theta \):

\[ \cos \theta = \frac{-1}{3} \]

### Step 5: Find \( \theta \)

To find \( \theta \), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]

This gives us the angle \( \theta \) between the two vectors.

### Final Answer

\[ \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]

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