If `vecA 2 hati +hatj -hatk, vecB=hati +2hatj +3hatk, vecC=6hati -2hatj-6hatk` angle between `(vecA+vecB) and vecC` will be

Options

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

Answer

To find the angle between the vectors \(\vec{A} + \vec{B}\) and \(\vec{C}\), we will 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 \(\vec{A}\) and \(\vec{B}\):
\[ \vec{A} + \vec{B} = (2\hat{i} + \hat{j} - \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) \]
Combine the components:
\[ = (2 + 1)\hat{i} + (1 + 2)\hat{j} + (-1 + 3)\hat{k} \]
\[ = 3\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 2: Calculate the Magnitude of \(\vec{A} + \vec{B}\)

The magnitude of a vector \(\vec{V} = x\hat{i} + y\hat{j} + z\hat{k}\) is given by:
\[ |\vec{V}| = \sqrt{x^2 + y^2 + z^2} \]

For \(\vec{A} + \vec{B} = 3\hat{i} + 3\hat{j} + 2\hat{k}\):
\[ |\vec{A} + \vec{B}| = \sqrt{3^2 + 3^2 + 2^2} = \sqrt{9 + 9 + 4} = \sqrt{22} \]

### Step 3: Calculate the Magnitude of \(\vec{C}\)

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

Now, calculate its magnitude:
\[ |\vec{C}| = \sqrt{6^2 + (-2)^2 + (-6)^2} = \sqrt{36 + 4 + 36} = \sqrt{76} \]

### Step 4: Calculate the Dot Product \((\vec{A} + \vec{B}) \cdot \vec{C}\)

The dot product of two vectors \(\vec{X} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}\) and \(\vec{Y} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}\) is given by:
\[ \vec{X} \cdot \vec{Y} = x_1 x_2 + y_1 y_2 + z_1 z_2 \]

For \(\vec{A} + \vec{B} = 3\hat{i} + 3\hat{j} + 2\hat{k}\) and \(\vec{C} = 6\hat{i} - 2\hat{j} - 6\hat{k}\):
\[ (\vec{A} + \vec{B}) \cdot \vec{C} = (3)(6) + (3)(-2) + (2)(-6) \]
\[ = 18 - 6 - 12 = 0 \]

### Step 5: Calculate the Angle \(\theta\)

The cosine of the angle \(\theta\) between two vectors is given by:
\[ \cos \theta = \frac{(\vec{A} + \vec{B}) \cdot \vec{C}}{|\vec{A} + \vec{B}| |\vec{C}|} \]

Substituting the values we found:
\[ \cos \theta = \frac{0}{|\vec{A} + \vec{B}| |\vec{C}|} = 0 \]

Since \(\cos \theta = 0\), we find that:
\[ \theta = 90^\circ \]

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

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