If ` veca , vecb` are unit vectors such that the vector ` veca + 3vecb ` is peependicular to ` 7 veca - vecb and veca -4vecb` is prependicular to ` 7 veca -2vecb` then the angle between ` veca and vecb` is

Options

A. ` pi//6`
B. ` pi//4`
C. ` pi//3`
D. ` pi//2`

Explanation

To solve the problem step by step, we will use the properties of dot products and the fact that the vectors are unit vectors.

### Step 1: Set Up the Problem
Given that \( \vec{a} \) and \( \vec{b} \) are unit vectors, we know:
- \( |\vec{a}| = 1 \)
- \( |\vec{b}| = 1 \)

We have two conditions:
1. \( \vec{a} + 3\vec{b} \) is perpendicular to \( 7\vec{a} - 5\vec{b} \)
2. \( \vec{a} - 4\vec{b} \) is perpendicular to \( 7\vec{a} - 2\vec{b} \)

### Step 2: Use the Perpendicular Condition
For two vectors \( \vec{u} \) and \( \vec{v} \) to be perpendicular, their dot product must equal zero:
\[ \vec{u} \cdot \vec{v} = 0 \]

#### Condition 1:
Set \( \vec{u} = \vec{a} + 3\vec{b} \) and \( \vec{v} = 7\vec{a} - 5\vec{b} \):
\[ (\vec{a} + 3\vec{b}) \cdot (7\vec{a} - 5\vec{b}) = 0 \]

Expanding this:
\[ \vec{a} \cdot (7\vec{a}) + 3\vec{b} \cdot (7\vec{a}) - 5\vec{a} \cdot \vec{b} - 15\vec{b} \cdot \vec{b} = 0 \]
\[ 7|\vec{a}|^2 + 21(\vec{a} \cdot \vec{b}) - 5(\vec{a} \cdot \vec{b}) - 15|\vec{b}|^2 = 0 \]
Since \( |\vec{a}|^2 = 1 \) and \( |\vec{b}|^2 = 1 \):
\[ 7 + 16(\vec{a} \cdot \vec{b}) - 15 = 0 \]
\[ 16(\vec{a} \cdot \vec{b}) = 8 \]
\[ \vec{a} \cdot \vec{b} = \frac{1}{2} \]

### Step 3: Find the Angle from the Dot Product
Using the dot product formula:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the values:
\[ \frac{1}{2} = 1 \cdot 1 \cdot \cos \theta \]
\[ \cos \theta = \frac{1}{2} \]
This implies:
\[ \theta = \frac{\pi}{3} \]

### Step 4: Verify with the Second Condition
Now we check the second condition:
Set \( \vec{u} = \vec{a} - 4\vec{b} \) and \( \vec{v} = 7\vec{a} - 2\vec{b} \):
\[ (\vec{a} - 4\vec{b}) \cdot (7\vec{a} - 2\vec{b}) = 0 \]

Expanding this:
\[ \vec{a} \cdot (7\vec{a}) - 2\vec{a} \cdot \vec{b} - 28\vec{b} \cdot \vec{a} + 8\vec{b} \cdot \vec{b} = 0 \]
\[ 7|\vec{a}|^2 - 30(\vec{a} \cdot \vec{b}) + 8|\vec{b}|^2 = 0 \]
Substituting \( |\vec{a}|^2 = 1 \) and \( |\vec{b}|^2 = 1 \):
\[ 7 - 30(\vec{a} \cdot \vec{b}) + 8 = 0 \]
\[ 15 - 30(\vec{a} \cdot \vec{b}) = 0 \]
\[ 30(\vec{a} \cdot \vec{b}) = 15 \]
\[ \vec{a} \cdot \vec{b} = \frac{1}{2} \]
This confirms our previous result.

### Final Answer
Thus, the angle between \( \vec{a} \) and \( \vec{b} \) is:
\[ \theta = \frac{\pi}{3} \]

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