Vectors a and b are such that `|a|=1,|b|=4 and a.b=2." If " c=2axxb-3b`, then the angle between b and c is

Options

A. `(pi)/(6)`
B. `(5pi)/(6)`
C. `(pi)/(3)`
D. `(2pi)/(3)`

Explanation

To solve the problem step by step, we need to find the angle between the vectors \( \mathbf{b} \) and \( \mathbf{c} \), where \( \mathbf{c} = 2 \mathbf{a} \times \mathbf{b} - 3 \mathbf{b} \).

### Step 1: Given Information
We have:
- \( |\mathbf{a}| = 1 \)
- \( |\mathbf{b}| = 4 \)
- \( \mathbf{a} \cdot \mathbf{b} = 2 \)

### Step 2: Find the Angle \( \theta \) Between \( \mathbf{a} \) and \( \mathbf{b} \)
Using the dot product formula:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]
Substituting the known values:
\[ 2 = 1 \cdot 4 \cdot \cos \theta \]
\[ \cos \theta = \frac{2}{4} = \frac{1}{2} \]
Thus, \( \theta = 60^\circ \).

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

### Step 4: Calculate \( |\mathbf{a} \times \mathbf{b}|^2 \)
\[ |\mathbf{a} \times \mathbf{b}|^2 = (2\sqrt{3})^2 = 12 \]

### Step 5: Find \( |\mathbf{c}|^2 \)
Using the expression for \( \mathbf{c} \):
\[ \mathbf{c} = 2(\mathbf{a} \times \mathbf{b}) - 3\mathbf{b} \]
Calculating \( |\mathbf{c}|^2 \):
\[ |\mathbf{c}|^2 = |2(\mathbf{a} \times \mathbf{b}) - 3\mathbf{b}|^2 \]
Using the formula \( |\mathbf{x} - \mathbf{y}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y} \):
\[ |\mathbf{c}|^2 = |2(\mathbf{a} \times \mathbf{b})|^2 + |-3\mathbf{b}|^2 - 2(2(\mathbf{a} \times \mathbf{b})) \cdot (-3\mathbf{b}) \]
Calculating each term:
\[ |2(\mathbf{a} \times \mathbf{b})|^2 = 4 \cdot 12 = 48 \]
\[ |-3\mathbf{b}|^2 = 9 \cdot 16 = 144 \]
The dot product \( 2(\mathbf{a} \times \mathbf{b}) \cdot (-3\mathbf{b}) = 0 \) because \( \mathbf{a} \times \mathbf{b} \) is perpendicular to \( \mathbf{b} \).

Thus:
\[ |\mathbf{c}|^2 = 48 + 144 = 192 \]

### Step 6: Find the Angle \( \phi \) Between \( \mathbf{b} \) and \( \mathbf{c} \)
Using the formula:
\[ \mathbf{b} \cdot \mathbf{c} = |\mathbf{b}| |\mathbf{c}| \cos \phi \]
Calculating \( \mathbf{b} \cdot \mathbf{c} \):
\[ \mathbf{b} \cdot \mathbf{c} = \mathbf{b} \cdot (2(\mathbf{a} \times \mathbf{b}) - 3\mathbf{b}) = 2\mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) - 3|\mathbf{b}|^2 \]
The first term is zero (as \( \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = 0 \)):
\[ \mathbf{b} \cdot \mathbf{c} = -3 \cdot 16 = -48 \]

Now substituting into the angle formula:
\[ -48 = 4 \cdot \sqrt{192} \cos \phi \]
Calculating \( \sqrt{192} = 8\sqrt{3} \):
\[ -48 = 4 \cdot 8\sqrt{3} \cos \phi \]
\[ -48 = 32\sqrt{3} \cos \phi \]
\[ \cos \phi = -\frac{48}{32\sqrt{3}} = -\frac{3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2} \]

### Final Step: Determine \( \phi \)
The angle \( \phi \) corresponds to:
\[ \phi = 150^\circ \quad \text{(since } \cos(150^\circ) = -\frac{\sqrt{3}}{2}\text{)} \]

Thus, the angle between \( \mathbf{b} \) and \( \mathbf{c} \) is \( 150^\circ \).