Question

Let $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}| = 1$, $|\bar{b}| = 4$, $\bar{a} \cdot \bar{b} = 2$. If $\bar{c} = (2\bar{a} \times \bar{b}) - 3\bar{b}$, then the angle between $\bar{b}$ and $\bar{c}$ is

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{6}$
• C. $\frac{3\pi}{4}$
• D. $\frac{5\pi}{6}$

Hint:

Vector $\bar{x} \times \bar{y}$ is always perpendicular to both $\bar{x}$ and $\bar{y}$. This makes many dot products zero!

Answer 1

The Correct Option is D

Step 1: Concept
The angle $\theta$ between two vectors $\bar{b}$ and $\bar{c}$ is found using the dot product formula $\cos \theta = \frac{\bar{b} \cdot \bar{c}}{|\bar{b}| |\bar{c}|}$.

Step 2: Meaning
Recall that any vector $\bar{b}$ is orthogonal to its cross product with another vector, so $\bar{b} \cdot (\bar{a} \times \bar{b}) = 0$.

Step 3: Analysis
$\bar{b} \cdot \bar{c} = \bar{b} \cdot (2\bar{a} \times \bar{b} - 3\bar{b}) = 2\bar{b} \cdot (\bar{a} \times \bar{b}) - 3\bar{b} \cdot \bar{b} = 0 - 3|\bar{b}|^2 = -3(16) = -48$. Now find $|\bar{c}|$: Since $(2\bar{a} \times \bar{b})$ is perpendicular to $3\bar{b}$, we use Pythagoras: $|\bar{c}|^2 = |2\bar{a} \times \bar{b}|^2 + |-3\bar{b}|^2$. $|\bar{a} \times \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2 = (1)(16) - 4 = 12$. $|\bar{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$. $|\bar{c}| = \sqrt{192} = 8\sqrt{3}$. $\cos \theta = \frac{-48}{4 \cdot 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$.

Step 4: Conclusion
$\theta = \cos^{-1}(-\sqrt{3}/2) = \frac{5\pi}{6}$. Final Answer: (D)