`veca and vecb` are two vectors such that `|veca|=1 ,|vecb|=4 and veca.` Vecb `=2 . If `vecc` = (2vecaxx vecb) - 3vecb` then find angle between `vecb and vecc`.

Options

A. A`pi/3`
B. B`pi/6`
C. C`(3pi)/4`
D. D`(5pi)/6`

Explanation

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

### Step-by-Step Solution:

1. **Given Information**:
- \(|\vec{a}| = 1\)
- \(|\vec{b}| = 4\)
- \(\vec{a} \cdot \vec{b} = 2\)

2. **Using the Dot Product**:
The dot product of two vectors can be expressed as:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). Substituting the known values:
\[ 2 = 1 \cdot 4 \cdot \cos \theta \]
Simplifying gives:
\[ 2 = 4 \cos \theta \implies \cos \theta = \frac{1}{2} \]
Therefore, \(\theta = 60^\circ\).

3. **Finding \(\vec{c}\)**:
The vector \(\vec{c}\) is defined as:
\[ \vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b} \]

4. **Calculating \(|\vec{c}|\)**:
First, we need to find \(|\vec{a} \times \vec{b}|\):
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
We already know \(|\vec{a}| = 1\), \(|\vec{b}| = 4\), and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\[ |\vec{a} \times \vec{b}| = 1 \cdot 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

5. **Substituting into \(\vec{c}\)**:
Now, substituting back into the expression for \(\vec{c}\):
\[ |\vec{c}| = |2(\vec{a} \times \vec{b}) - 3\vec{b}| \]
We can find the magnitude of \(\vec{c}\) using the formula for the magnitude of a vector subtraction:
\[ |\vec{c}|^2 = |2(\vec{a} \times \vec{b})|^2 + |-3\vec{b}|^2 - 2|2(\vec{a} \times \vec{b})||-3\vec{b}|\cos \phi \]
where \(\phi\) is the angle between \(\vec{a} \times \vec{b}\) and \(\vec{b}\). Since \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), \(\phi = 90^\circ\) and \(\cos \phi = 0\):
\[ |\vec{c}|^2 = |2(2\sqrt{3})|^2 + |-3(4)|^2 \]
\[ = 4(3) + 9(16) = 12 + 144 = 156 \]
Therefore, \(|\vec{c}| = \sqrt{156} = 2\sqrt{39}\).

6. **Finding the Angle Between \(\vec{b}\) and \(\vec{c}\)**:
Now, we can find the angle \(\alpha\) between \(\vec{b}\) and \(\vec{c}\) using the dot product:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \alpha \]
We already know \(|\vec{b}| = 4\) and \(|\vec{c}| = 2\sqrt{39}\). Now we need to find \(\vec{b} \cdot \vec{c}\):
\[ \vec{b} \cdot \vec{c} = 2(\vec{a} \cdot \vec{b}) - 3|\vec{b}|^2 \]
\[ = 2(2) - 3(16) = 4 - 48 = -44 \]

7. **Substituting Values**:
Now substituting into the dot product equation:
\[ -44 = 4 \cdot 2\sqrt{39} \cos \alpha \]
\[ -44 = 8\sqrt{39} \cos \alpha \implies \cos \alpha = \frac{-44}{8\sqrt{39}} = \frac{-11}{2\sqrt{39}} \]

8. **Finding \(\alpha\)**:
The angle \(\alpha\) can be found using the inverse cosine function:
\[ \alpha = \cos^{-1}\left(\frac{-11}{2\sqrt{39}}\right) \]

### Final Answer:
The angle between \(\vec{b}\) and \(\vec{c}\) is given by:
\[ \alpha = \cos^{-1}\left(\frac{-11}{2\sqrt{39}}\right) \]