Let `2veca = vecb xx vecc + 2vecb` where `veca, vecb and vecc` are three unit vectors, then sum of all possible values of `|3veca + 4vecb + 5vecc|` is

Options

A. 10
B. 12
C. 14
D. 16

Answer

To solve the problem, we start with the given equation involving the unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\):

\[ 2\vec{a} = \vec{b} \times \vec{c} + 2\vec{b} \]

### Step 1: Dot the equation with \(\vec{b}\)

We will first take the dot product of both sides of the equation with \(\vec{b}\):

\[ \vec{b} \cdot (2\vec{a}) = \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (2\vec{b}) \]

### Step 2: Simplify the left-hand side

The left-hand side simplifies to:

\[ 2(\vec{a} \cdot \vec{b}) \]

### Step 3: Simplify the right-hand side

The term \(\vec{b} \cdot (\vec{b} \times \vec{c})\) is zero because the dot product of a vector with a vector perpendicular to it is zero. The second term simplifies to:

\[ 2(\vec{b} \cdot \vec{b}) = 2(1) = 2 \]

So, we have:

\[ 2(\vec{a} \cdot \vec{b}) = 2 \]

### Step 4: Solve for \(\vec{a} \cdot \vec{b}\)

Dividing both sides by 2 gives us:

\[ \vec{a} \cdot \vec{b} = 1 \]

### Step 5: Interpret the result

Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\) implies:

\[ 1 = 1 \cdot 1 \cdot \cos \theta \implies \cos \theta = 1 \]

This means that the angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) is \(0^\circ\), so:

\[ \vec{a} = \vec{b} \]

### Step 6: Substitute back into the original equation

Substituting \(\vec{a} = \vec{b}\) into the original equation:

\[ 2\vec{a} = \vec{a} \times \vec{c} + 2\vec{a} \]

This simplifies to:

\[ 0 = \vec{a} \times \vec{c} \]

### Step 7: Analyze the cross product

The equation \(\vec{a} \times \vec{c} = 0\) implies that \(\vec{a}\) and \(\vec{c}\) are parallel. Thus, we have two cases:

1. \(\vec{c} = \vec{a}\)
2. \(\vec{c} = -\vec{a}\)

### Step 8: Calculate \(|3\vec{a} + 4\vec{b} + 5\vec{c}|\)

#### Case 1: \(\vec{c} = \vec{a}\)

\[ |3\vec{a} + 4\vec{a} + 5\vec{a}| = |(3 + 4 + 5)\vec{a}| = |12\vec{a}| = 12 \]

#### Case 2: \(\vec{c} = -\vec{a}\)

\[ |3\vec{a} + 4\vec{a} + 5(-\vec{a})| = |(3 + 4 - 5)\vec{a}| = |2\vec{a}| = 2 \]

### Step 9: Sum of all possible values

The possible values of \(|3\vec{a} + 4\vec{b} + 5\vec{c}|\) are \(12\) and \(2\). Therefore, the sum of all possible values is:

\[ 12 + 2 = 14 \]

### Final Answer

The sum of all possible values of \(|3\vec{a} + 4\vec{b} + 5\vec{c}|\) is:

\[ \boxed{14} \]