`bara` & `barb` be two vectors such that `|a|=1,|b|=4`& `bara.barb=2`. If `barc=(2baraxxbarb)-3barb`, then which of the following is/are correct?

Options

A. `barb.barc=48`
B. `barb.barc=-48`
C. Angle between `barb` & `barc` is`(5pi)/6`
D. Angle between `barb` & `barc` is `(pi)/6`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given two vectors \( \mathbf{a} \) and \( \mathbf{b} \) such that:
- \( |\mathbf{a}| = 1 \)
- \( |\mathbf{b}| = 4 \)
- \( \mathbf{a} \cdot \mathbf{b} = 2 \)

We also have \( \mathbf{c} = 2(\mathbf{a} \times \mathbf{b}) - 3\mathbf{b} \).

### Step 2: Find the Angle Between the Vectors
Using the dot product formula:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]
Substituting the known values:
\[ 2 = 1 \cdot 4 \cdot \cos \theta \]
This simplifies to:
\[ \cos \theta = \frac{2}{4} = \frac{1}{2} \]
Thus, the angle \( \theta \) is:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Step 3: Calculate the Cross Product Magnitude
Using the sine formula for the cross product:
\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \theta \]
Substituting the values:
\[ |\mathbf{a} \times \mathbf{b}| = 1 \cdot 4 \cdot \sin(60^\circ) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

### Step 4: Substitute into the Expression for \( \mathbf{c} \)
Now substituting into the expression for \( \mathbf{c} \):
\[ \mathbf{c} = 2(\mathbf{a} \times \mathbf{b}) - 3\mathbf{b} \]
Calculating \( 2(\mathbf{a} \times \mathbf{b}) \):
\[ 2(\mathbf{a} \times \mathbf{b}) = 2(2\sqrt{3}) = 4\sqrt{3} \]
Thus:
\[ \mathbf{c} = 4\sqrt{3} - 3\mathbf{b} \]

### Step 5: Calculate the Magnitude of \( \mathbf{c} \)
To find the magnitude of \( \mathbf{c} \), we need to calculate:
\[ |\mathbf{c}|^2 = |4\sqrt{3} - 3\mathbf{b}|^2 \]
Using the formula for the magnitude:
\[ |\mathbf{c}|^2 = |4\sqrt{3}|^2 + |3\mathbf{b}|^2 - 2(4\sqrt{3})(3|\mathbf{b}|)\cos(180^\circ) \]
Substituting \( |\mathbf{b}| = 4 \):
\[ |\mathbf{c}|^2 = (4\sqrt{3})^2 + (3 \cdot 4)^2 + 2(4\sqrt{3})(3 \cdot 4) \]
Calculating each term:
\[ = 48 + 144 + 2(4\sqrt{3})(12) \]
\[ = 48 + 144 - 96\sqrt{3} \]
Thus:
\[ |\mathbf{c}|^2 = 192 \]
Taking the square root gives:
\[ |\mathbf{c}| = 8\sqrt{3} \]

### Step 6: Find the Angle Between \( \mathbf{b} \) and \( \mathbf{c} \)
Now we find the angle between \( \mathbf{b} \) and \( \mathbf{c} \) using the dot product:
\[ \mathbf{c} \cdot \mathbf{b} = 2(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} - 3|\mathbf{b}|^2 \]
Since \( \mathbf{a} \times \mathbf{b} \) is perpendicular to \( \mathbf{b} \), the first term is zero:
\[ \mathbf{c} \cdot \mathbf{b} = -3(4^2) = -48 \]

Using the cosine formula:
\[ \cos \alpha = \frac{\mathbf{c} \cdot \mathbf{b}}{|\mathbf{c}| |\mathbf{b}|} \]
Substituting the values:
\[ \cos \alpha = \frac{-48}{(8\sqrt{3})(4)} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2} \]
Thus:
\[ \alpha = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} \]

### Conclusion
The angle between \( \mathbf{b} \) and \( \mathbf{c} \) is \( \frac{5\pi}{6} \).