Find the angle between `vec(A) = hat(i) + 2hat(j) - hat(k) and vec(B) = - hat(i) + hat(j) - 2hat(k)`

Explanation

To find the angle between the vectors \(\vec{A} = \hat{i} + 2\hat{j} - \hat{k}\) and \(\vec{B} = -\hat{i} + \hat{j} - 2\hat{k}\), we can use the formula for the dot product of two vectors:

\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]

where \(\theta\) is the angle between the vectors.

### Step 1: Calculate the dot product \(\vec{A} \cdot \vec{B}\)

\[ \vec{A} \cdot \vec{B} = (1)(-1) + (2)(1) + (-1)(-2) \]
\[ = -1 + 2 + 2 = 3 \]

### Step 2: Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)

For \(\vec{A}\):
\[ |\vec{A}| = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

For \(\vec{B}\):
\[ |\vec{B}| = \sqrt{(-1)^2 + (1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Step 3: Substitute into the cosine formula

Using the dot product and the magnitudes calculated:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2} \]

### Step 4: Find the angle \(\theta\)

To find \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]
\[ \theta = 60^\circ \]

### Final Answer
The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(60^\circ\).

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