Find the angle between `(vec(a)+vec(b))` and `(vec(a)-vec(b))`, if `vec(a)=(2 hat(i)-hat(j)+3hat(k))` and `vec(b)=(3hat(i)+hat(j)+2hat(k))`.

Explanation

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

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

### Step 2: Calculate \( \vec{c} = \vec{a} + \vec{b} \)
\[ \vec{c} = (2\hat{i} - \hat{j} + 3\hat{k}) + (3\hat{i} + \hat{j} + 2\hat{k}) \]
Combine the components:
\[ \vec{c} = (2 + 3)\hat{i} + (-1 + 1)\hat{j} + (3 + 2)\hat{k} \]
\[ \vec{c} = 5\hat{i} + 0\hat{j} + 5\hat{k} = 5\hat{i} + 5\hat{k} \]

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

### Step 4: Find the dot product \( \vec{c} \cdot \vec{d} \)
\[ \vec{c} \cdot \vec{d} = (5\hat{i} + 5\hat{k}) \cdot (-1\hat{i} - 2\hat{j} + 1\hat{k}) \]
Calculating the dot product:
\[ = 5(-1) + 0(-2) + 5(1) = -5 + 0 + 5 = 0 \]

### Step 5: Calculate the magnitudes of \( \vec{c} \) and \( \vec{d} \)
\[ |\vec{c}| = \sqrt{(5)^2 + (0)^2 + (5)^2} = \sqrt{25 + 0 + 25} = \sqrt{50} = 5\sqrt{2} \]
\[ |\vec{d}| = \sqrt{(-1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

### Step 6: Use the cosine formula to find the angle \( \theta \)
Using the formula:
\[ \cos \theta = \frac{\vec{c} \cdot \vec{d}}{|\vec{c}| |\vec{d}|} \]
Substituting the values:
\[ \cos \theta = \frac{0}{(5\sqrt{2})(\sqrt{6})} = 0 \]

### Step 7: Determine the angle \( \theta \)
Since \( \cos \theta = 0 \):
\[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{or} \quad 90^\circ \]

Thus, the angle between \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) is \( 90^\circ \).

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