If `vec(A) = 2 I + j - k , vec(B) = I + 2 j + 3 k` , and `vec(C ) = 6 i - 2 j - 6 k`,then the angle between `(vec(A) + vec(B))` and `vec(C )` will be

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

To find the angle between the vectors \(\vec{A} + \vec{B}\) and \(\vec{C}\), we can follow these steps:

### Step 1: Calculate \(\vec{A} + \vec{B}\)

Given:
\[ \vec{A} = 2\hat{i} + \hat{j} - \hat{k} \]
\[ \vec{B} = \hat{i} + 2\hat{j} + 3\hat{k} \]

Now, we add the vectors:
\[ \vec{A} + \vec{B} = (2\hat{i} + \hat{j} - \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) \]

Combine the components:
- For \(\hat{i}\): \(2 + 1 = 3\)
- For \(\hat{j}\): \(1 + 2 = 3\)
- For \(\hat{k}\): \(-1 + 3 = 2\)

Thus,
\[ \vec{A} + \vec{B} = 3\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 2: Identify \(\vec{C}\)

Given:
\[ \vec{C} = 6\hat{i} - 2\hat{j} - 6\hat{k} \]

### Step 3: Calculate the dot product \((\vec{A} + \vec{B}) \cdot \vec{C}\)

Using the vectors we calculated:
\[ \vec{R} = \vec{A} + \vec{B} = 3\hat{i} + 3\hat{j} + 2\hat{k} \]

Now, calculate the dot product:
\[ \vec{R} \cdot \vec{C} = (3\hat{i} + 3\hat{j} + 2\hat{k}) \cdot (6\hat{i} - 2\hat{j} - 6\hat{k}) \]

Calculating each component:
- For \(\hat{i}\): \(3 \cdot 6 = 18\)
- For \(\hat{j}\): \(3 \cdot (-2) = -6\)
- For \(\hat{k}\): \(2 \cdot (-6) = -12\)

Thus,
\[ \vec{R} \cdot \vec{C} = 18 - 6 - 12 = 0 \]

### Step 4: Calculate the magnitudes of \(\vec{R}\) and \(\vec{C}\)

Magnitude of \(\vec{R}\):
\[ |\vec{R}| = \sqrt{(3^2 + 3^2 + 2^2)} = \sqrt{9 + 9 + 4} = \sqrt{22} \]

Magnitude of \(\vec{C}\):
\[ |\vec{C}| = \sqrt{(6^2 + (-2)^2 + (-6)^2)} = \sqrt{36 + 4 + 36} = \sqrt{76} \]

### Step 5: Use the dot product to find the angle \(\theta\)

Using the formula:
\[ \vec{R} \cdot \vec{C} = |\vec{R}| |\vec{C}| \cos \theta \]

Since \(\vec{R} \cdot \vec{C} = 0\):
\[ 0 = |\vec{R}| |\vec{C}| \cos \theta \]

This implies:
\[ \cos \theta = 0 \]

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

The angle \(\theta\) for which \(\cos \theta = 0\) is:
\[ \theta = 90^\circ \]

### Final Answer:
The angle between \((\vec{A} + \vec{B})\) and \(\vec{C}\) is \(90^\circ\).

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