If `vecA = hati + 2hatj - hatk , vecB = - hati + hatj - 2hatk ` , then angle between `vecA and vecB` is

Options

A. `pi/2`
B. 0
C. `pi`
D. `pi/3`

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 will use the dot product formula. The dot product of two vectors can be expressed in terms of the angle between them as follows:

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

Where:
- \(\vec{A} \cdot \vec{B}\) is the dot product of vectors \(\vec{A}\) and \(\vec{B}\).
- \(|\vec{A}|\) and \(|\vec{B}|\) are the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\) respectively.
- \(\theta\) is the angle between the two vectors.

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

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

Calculating each term:

\[ = -1 + 2 + 2 = 3 \]

### Step 2: Calculate the magnitude of \(\vec{A}\)

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

### Step 3: Calculate the magnitude of \(\vec{B}\)

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

### Step 4: Substitute the values into the dot product formula

Using the values calculated:

\[ 3 = |\vec{A}| |\vec{B}| \cos \theta \]

Substituting the magnitudes:

\[ 3 = \sqrt{6} \cdot \sqrt{6} \cdot \cos \theta \]

\[ 3 = 6 \cos \theta \]

### Step 5: Solve for \(\cos \theta\)

\[ \cos \theta = \frac{3}{6} = \frac{1}{2} \]

### Step 6: Find the angle \(\theta\)

The angle whose cosine is \(\frac{1}{2}\) is:

\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \text{ radians} = 60^\circ \]

### Final Answer

The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(60^\circ\) or \(\frac{\pi}{3}\) radians.

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