Question

Let \( \vec{a} \) and \( \vec{b} \) be two vectors such that \( |\vec{a}| = 1 \), \( |\vec{b}| = 4 \) and \( \vec{a} \cdot \vec{b} = 2 \).If \( \vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b} \) and the angle between \( \vec{b} \) and \( \vec{c} \) is \( \alpha \), then \( 192 \sin^2 \alpha \) is equal to _____

Answer 1

Correct Answer: 48

Step 1: We start with the equation for the dot product of vectors \( \vec{b} \) and \( \vec{c} \): 

\[ \vec{b} \cdot \vec{c} = (2\vec{a} \times \vec{b}) \cdot \vec{b} - 3|\vec{b}|^2 \]

Step 2: The magnitude of vector \( \vec{b} \) and the dot product of \( \vec{b} \) and \( \vec{c} \) are given as:

\[ |\vec{b}||\vec{c}| \cos \alpha = -3|\vec{b}|^2 \]

Substitute the known value of \( |\vec{b}| = 4 \):

\[ |\vec{b}| |\vec{c}| \cos \alpha = -12, \quad \text{where} \quad |\vec{b}| = 4 \]

Step 3: The dot product of vectors \( \vec{a} \) and \( \vec{b} \) is given:

\[ \vec{a} \cdot \vec{b} = 2 \]

Step 4: We now use the cosine identity to find \( \theta \):

\[ \cos \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \]

Step 5: The magnitude of \( \vec{c} \) is given as:

\[ |\vec{c}|^2 = |(2\vec{a} \times \vec{b}) - 3\vec{b}|^2 \]

We calculate each term of this expression:

\[ = 64 \times \frac{3}{4} + 144 = 192 \]

Step 6: Substituting in \( \cos^2 \alpha \), we get:

\[ |\vec{c}|^2 \cos^2 \alpha = 144 \]

Step 7: Now, solving for \( \sin^2 \alpha \), we get:

\[ 192 \cos^2 \alpha = 144 \quad \Rightarrow \quad 192 \sin^2 \alpha = 48 \]

Answer 2

Given:

\[ \vec{b} \cdot \vec{c} = \left( 2\vec{a} \times \vec{b} \right) \cdot \vec{b} - 3|\vec{b}|^2 \]

We know:

\[ |\vec{b}||\vec{c}| \cos \alpha = -12, \quad \text{as } |\vec{b}| = 4, \; \vec{a} \cdot \vec{b} = 2 \]

Calculate:

\[ \cos \alpha = \frac{1}{2}, \quad \alpha = \frac{\pi}{3} \]

Now:

\[ |\vec{c}|^2 = |(2\vec{a} \times \vec{b}) - 3\vec{b}|^2 = 64 \times \frac{3}{4} + 144 = 192 \]

Therefore:

\[ 192 \cos^2 \alpha = 144, \quad 192 \sin^2 \alpha = 48 \]