If `vec(a)=(hat(i)+2hat(j)-3hat(k)) and vec(b)=(3hat(i)-hat(j)+2hat(k))` then the angle between `(vec(a)+vec(b))` and `(vec(a)-vec(b))` is

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(2)`
D. `(2pi)/(3)`

Explanation

To find the angle between the vectors \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\), we can follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \]
\[ \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \]

### Step 2: Calculate \(\vec{a} + \vec{b}\)
\[ \vec{a} + \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) + (3\hat{i} - \hat{j} + 2\hat{k}) \]
Combine the components:
\[ = (1 + 3)\hat{i} + (2 - 1)\hat{j} + (-3 + 2)\hat{k} \]
\[ = 4\hat{i} + 1\hat{j} - 1\hat{k} \]
Thus,
\[ \vec{a} + \vec{b} = 4\hat{i} + \hat{j} - \hat{k} \]

### Step 3: Calculate \(\vec{a} - \vec{b}\)
\[ \vec{a} - \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) - (3\hat{i} - \hat{j} + 2\hat{k}) \]
Combine the components:
\[ = (1 - 3)\hat{i} + (2 + 1)\hat{j} + (-3 - 2)\hat{k} \]
\[ = -2\hat{i} + 3\hat{j} - 5\hat{k} \]
Thus,
\[ \vec{a} - \vec{b} = -2\hat{i} + 3\hat{j} - 5\hat{k} \]

### Step 4: Calculate the dot product \((\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})\)
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = (4\hat{i} + \hat{j} - \hat{k}) \cdot (-2\hat{i} + 3\hat{j} - 5\hat{k}) \]
Calculating the dot product:
\[ = 4(-2) + 1(3) + (-1)(-5) \]
\[ = -8 + 3 + 5 \]
\[ = 0 \]

### Step 5: Calculate the magnitudes
1. Magnitude of \(\vec{a} + \vec{b}\):
\[ |\vec{a} + \vec{b}| = \sqrt{(4)^2 + (1)^2 + (-1)^2} = \sqrt{16 + 1 + 1} = \sqrt{18} \]

2. Magnitude of \(\vec{a} - \vec{b}\):
\[ |\vec{a} - \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (-5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38} \]

### Step 6: Use the formula for the angle \(\theta\)
The cosine of the angle \(\theta\) is given by:
\[ \cos(\theta) = \frac{(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})}{|\vec{a} + \vec{b}| \cdot |\vec{a} - \vec{b}|} \]
Substituting the values:
\[ \cos(\theta) = \frac{0}{\sqrt{18} \cdot \sqrt{38}} = 0 \]

### Step 7: Determine the angle
Since \(\cos(\theta) = 0\), we have:
\[ \theta = 90^\circ \]

### Final Answer
The angle between \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\) is \(90^\circ\).

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