The angle between `a+2b" and "a-3b` if `abs(a)=1, abs(b)=2` and angle between a and b is `60^(@)` is

Options

A. an acute angle
B. `cos^(-1)""(-24)/(sqrt(21)sqrt(31))`
C. `cos^(-1)""(24)/(sqrt(21)sqrt(31))`
D. `cos^(-1)""-1/3`

Answer

To find the angle between the vectors \( \mathbf{a} + 2\mathbf{b} \) and \( \mathbf{a} - 3\mathbf{b} \), given that \( |\mathbf{a}| = 1 \), \( |\mathbf{b}| = 2 \), and the angle between \( \mathbf{a} \) and \( \mathbf{b} \) is \( 60^\circ \), we can follow these steps:

### Step 1: Calculate the dot product of the vectors
We start by calculating the dot product of the vectors \( \mathbf{x} = \mathbf{a} + 2\mathbf{b} \) and \( \mathbf{y} = \mathbf{a} - 3\mathbf{b} \).

\[ \mathbf{x} \cdot \mathbf{y} = (\mathbf{a} + 2\mathbf{b}) \cdot (\mathbf{a} - 3\mathbf{b}) \]

Using the distributive property of the dot product, we have:

\[ \mathbf{x} \cdot \mathbf{y} = \mathbf{a} \cdot \mathbf{a} - 3\mathbf{a} \cdot \mathbf{b} + 2\mathbf{b} \cdot \mathbf{a} - 6\mathbf{b} \cdot \mathbf{b} \]

### Step 2: Substitute known values
We know that \( \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 = 1^2 = 1 \) and \( \mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2 = 2^2 = 4 \). The dot product \( \mathbf{a} \cdot \mathbf{b} \) can be calculated using the formula:

\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(60^\circ) = 1 \cdot 2 \cdot \frac{1}{2} = 1 \]

Now substituting these values into the dot product:

\[ \mathbf{x} \cdot \mathbf{y} = 1 - 3(1) + 2(1) - 6(4) \]
\[ = 1 - 3 + 2 - 24 = -24 \]

### Step 3: Calculate the magnitudes of the vectors
Next, we need to find the magnitudes of \( \mathbf{x} \) and \( \mathbf{y} \).

#### Magnitude of \( \mathbf{x} = \mathbf{a} + 2\mathbf{b} \):
\[ |\mathbf{x}|^2 = (\mathbf{a} + 2\mathbf{b}) \cdot (\mathbf{a} + 2\mathbf{b}) = \mathbf{a} \cdot \mathbf{a} + 4\mathbf{b} \cdot \mathbf{b} + 4\mathbf{a} \cdot \mathbf{b} \]
\[ = 1 + 4(4) + 4(1) = 1 + 16 + 4 = 21 \]
\[ |\mathbf{x}| = \sqrt{21} \]

#### Magnitude of \( \mathbf{y} = \mathbf{a} - 3\mathbf{b} \):
\[ |\mathbf{y}|^2 = (\mathbf{a} - 3\mathbf{b}) \cdot (\mathbf{a} - 3\mathbf{b}) = \mathbf{a} \cdot \mathbf{a} + 9\mathbf{b} \cdot \mathbf{b} - 6\mathbf{a} \cdot \mathbf{b} \]
\[ = 1 + 9(4) - 6(1) = 1 + 36 - 6 = 31 \]
\[ |\mathbf{y}| = \sqrt{31} \]

### Step 4: Use the cosine formula to find the angle
Now, we can use the formula for the cosine of the angle \( \theta \) between the two vectors:

\[ \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}| |\mathbf{y}|} \]
\[ \cos \theta = \frac{-24}{\sqrt{21} \cdot \sqrt{31}} = \frac{-24}{\sqrt{651}} \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{-24}{\sqrt{651}}\right) \]

This gives us the angle between the vectors \( \mathbf{a} + 2\mathbf{b} \) and \( \mathbf{a} - 3\mathbf{b} \).