Let `veca=x^(2)hati-3hatj+(x-3)hatk` and `vecb=hati+3hatj-(x-3)hatk` be two vectors such that `|veca|=|vecb|`. If angle between `4veca+7vecb and 7veca-4vecb` is equal to `theta`. Then `cos 2 theta` is equal to

Options

A. `-(1)/(2)`
B. `(sqrt3)/(2)`
C. `-1`
D. `(1)/(5)`

Answer

To solve the problem, we need to follow these steps:

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

### Step 2: Calculate the magnitudes of the vectors
The magnitudes of the vectors are given by:
\[ |\vec{a}| = \sqrt{(x^2)^2 + (-3)^2 + (x - 3)^2} \]
\[ |\vec{b}| = \sqrt{(1)^2 + (3)^2 + (-(x - 3))^2} \]

Calculating these:
\[ |\vec{a}| = \sqrt{x^4 + 9 + (x^2 - 6x + 9)} = \sqrt{x^4 + x^2 - 6x + 18} \]
\[ |\vec{b}| = \sqrt{1 + 9 + (x - 3)^2} = \sqrt{10 + (x^2 - 6x + 9)} = \sqrt{x^2 - 6x + 19} \]

### Step 3: Set the magnitudes equal
Since \( |\vec{a}| = |\vec{b}| \):
\[ \sqrt{x^4 + x^2 - 6x + 18} = \sqrt{x^2 - 6x + 19} \]
Squaring both sides:
\[ x^4 + x^2 - 6x + 18 = x^2 - 6x + 19 \]
Simplifying gives:
\[ x^4 + x^2 - 6x + 18 - x^2 + 6x - 19 = 0 \]
\[ x^4 - 1 = 0 \]
Factoring:
\[ (x^2 - 1)(x^2 + 1) = 0 \]
Thus, \( x^2 - 1 = 0 \) gives \( x = 1 \) or \( x = -1 \).

### Step 4: Find the angle between \( 4\vec{a} + 7\vec{b} \) and \( 7\vec{a} - 4\vec{b} \)
Let:
\[ \vec{u} = 4\vec{a} + 7\vec{b} \]
\[ \vec{v} = 7\vec{a} - 4\vec{b} \]

### Step 5: Calculate the dot product \( \vec{u} \cdot \vec{v} \)
Using the distributive property:
\[ \vec{u} \cdot \vec{v} = (4\vec{a} + 7\vec{b}) \cdot (7\vec{a} - 4\vec{b}) = 28\vec{a} \cdot \vec{a} - 16\vec{a} \cdot \vec{b} + 49\vec{b} \cdot \vec{b} - 28\vec{b} \cdot \vec{a} \]
Since \( \vec{a} \cdot \vec{b} = 0 \):
\[ \vec{u} \cdot \vec{v} = 28|\vec{a}|^2 + 49|\vec{b}|^2 \]

### Step 6: Calculate the magnitudes of \( \vec{u} \) and \( \vec{v} \)
\[ |\vec{u}| = \sqrt{(4|\vec{a}|)^2 + (7|\vec{b}|)^2} = \sqrt{16|\vec{a}|^2 + 49|\vec{b}|^2} \]
\[ |\vec{v}| = \sqrt{(7|\vec{a}|)^2 + (4|\vec{b}|)^2} = \sqrt{49|\vec{a}|^2 + 16|\vec{b}|^2} \]

### Step 7: Find \( \cos \theta \)
Using the formula:
\[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \]
Substituting the values:
\[ \cos \theta = \frac{28|\vec{a}|^2 + 49|\vec{b}|^2}{\sqrt{16|\vec{a}|^2 + 49|\vec{b}|^2} \cdot \sqrt{49|\vec{a}|^2 + 16|\vec{b}|^2}} \]

### Step 8: Calculate \( \cos 2\theta \)
Using the double angle formula:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]

### Final Answer
After simplifying, we find that \( \cos 2\theta = -1 \).