Three vectors `veca,vecb,vecc` are such that `vecaxxvecb= 3(vecaxxvecc)`Also`|veca|=|vecb|=1, |vecc|=1/3` If the angle between `vecb` and `vecc` is `60^@` then

Options

A. `b=3c+a`
B. `b=3c-a`
C. `a=6c+2b`
D. `a=6c-2b`

Answer

To solve the problem, we need to analyze the given information about the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Given:
1. \(\vec{a} \times \vec{b} = 3(\vec{a} \times \vec{c})\)
2. \(|\vec{a}| = |\vec{b}| = 1\)
3. \(|\vec{c}| = \frac{1}{3}\)
4. The angle between \(\vec{b}\) and \(\vec{c}\) is \(60^\circ\).

### Step 1: Write the Magnitude of the Cross Products
Using the formula for the magnitude of the cross product, we have:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\alpha) \]
\[ |\vec{a} \times \vec{c}| = |\vec{a}| |\vec{c}| \sin(\beta) \]
Where \(\alpha\) is the angle between \(\vec{a}\) and \(\vec{b}\), and \(\beta\) is the angle between \(\vec{a}\) and \(\vec{c}\).

### Step 2: Substitute the Magnitudes
Since \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\), we can simplify:
\[ |\vec{a} \times \vec{b}| = 1 \cdot 1 \cdot \sin(\alpha) = \sin(\alpha) \]
For \(\vec{c}\), we have \(|\vec{c}| = \frac{1}{3}\):
\[ |\vec{a} \times \vec{c}| = 1 \cdot \frac{1}{3} \cdot \sin(\beta) = \frac{1}{3} \sin(\beta) \]

### Step 3: Set Up the Equation
From the given condition:
\[ \sin(\alpha) = 3 \left(\frac{1}{3} \sin(\beta)\right) = \sin(\beta) \]
Thus, we have:
\[ \sin(\alpha) = \sin(\beta) \]

### Step 4: Conclude the Angles
Since \(\sin(\alpha) = \sin(\beta)\) and both angles are between \(0^\circ\) and \(90^\circ\), we can conclude:
\[ \alpha = \beta \]

### Step 5: Use the Cosine Rule
Now, we apply the cosine rule for the angles:
\[ \cos^2(\alpha) + \cos^2(\beta) + \cos^2(60^\circ) = 1 \]
Substituting \(\cos(60^\circ) = \frac{1}{2}\):
\[ \cos^2(\alpha) + \cos^2(\alpha) + \left(\frac{1}{2}\right)^2 = 1 \]
\[ 2\cos^2(\alpha) + \frac{1}{4} = 1 \]
\[ 2\cos^2(\alpha) = 1 - \frac{1}{4} = \frac{3}{4} \]
\[ \cos^2(\alpha) = \frac{3}{8} \]
Thus,
\[ \cos(\alpha) = \sqrt{\frac{3}{8}} = \frac{\sqrt{3}}{2\sqrt{2}} \]

### Step 6: Relate the Vectors
From the condition \(\vec{a} \times \vec{b} = 3(\vec{a} \times \vec{c})\), we can express this as:
\[ \vec{a} \times \vec{b} - 3(\vec{a} \times \vec{c}) = 0 \]
This implies that the vectors \(\vec{b}\) and \(3\vec{c}\) are parallel, leading to:
\[ \vec{b} = \vec{a} + 3\vec{c} \]

### Final Result
Thus, the correct relation is:
\[ \vec{b} = \vec{a} + 3\vec{c} \]