If `vecA=2hati-2hatj-hatk` and `vecB=hati+hatj`, then :
(a) Find angle between `vecA` and `vecB`.

Options

A. `vec(B)`
B. `vec(C)`
C. `vec(B). vec(C)`
D. `vec(B)xxvec(C)`

Explanation

To find the angle between the vectors \(\vec{A}\) and \(\vec{B}\), we can use the formula for the cosine of the angle \(\theta\) between two vectors:

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

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

Given:
\[ \vec{A} = 2\hat{i} - 2\hat{j} - \hat{k} \]
\[ \vec{B} = \hat{i} + \hat{j} \]

The dot product \(\vec{A} \cdot \vec{B}\) is calculated as follows:

\[ \vec{A} \cdot \vec{B} = (2\hat{i} - 2\hat{j} - \hat{k}) \cdot (\hat{i} + \hat{j}) \]

Calculating the dot product:

\[ = 2(\hat{i} \cdot \hat{i}) + (-2)(\hat{j} \cdot \hat{j}) + 0 \]
\[ = 2(1) + (-2)(1) + 0 = 2 - 2 + 0 = 0 \]

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

**Magnitude of \(\vec{A}\)**:
\[ |\vec{A}| = \sqrt{(2)^2 + (-2)^2 + (-1)^2} \]
\[ = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

**Magnitude of \(\vec{B}\)**:
\[ |\vec{B}| = \sqrt{(1)^2 + (1)^2} \]
\[ = \sqrt{1 + 1} = \sqrt{2} \]

### Step 3: Substitute into the cosine formula

Now substituting the dot product and magnitudes into the cosine formula:

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{0}{3 \cdot \sqrt{2}} = 0 \]

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

Since \(\cos \theta = 0\), we find:

\[ \theta = \cos^{-1}(0) = 90^\circ \]

### Final Answer

The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).

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