Question

Let \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \) be two vectors such that \( |\vec{a}| = 1 \), \( \vec{a} \times \vec{b} = 2 \), and \( |\vec{b}| = 4 \). If \( \vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b} \), then the angle between \( \vec{b} \) and \( \vec{c} \) is equal to:

• A. \( \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \)
• B. \( \cos^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)
• C. \( \cos^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
• D. \( \cos^{-1}\left(\frac{2}{3}\right) \)

Answer 1

The Correct Option is A

Given \( |\vec{a}| = 1 \), \( |\vec{b}| = 4 \), and \( \vec{a} \times \vec{b} = 2 \), we can find the magnitude of \( \vec{a} \times \vec{b} \) and then use it to determine \( \vec{c} \).

• Calculate \( |\vec{a} \times \vec{b}| \):

  \[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

• Find \( |\vec{c}| \) using \( \vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b} \):

  \[ |\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2 = 4(12) + 9(16) = 48 + 144 = 192 \]

  \[ |\vec{c}| = 8\sqrt{3} \]

• Find \( \cos \theta \) between \( \vec{b} \) and \( \vec{c} \):

  \[ \cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|} = \frac{-48}{4 \times 8\sqrt{3}} = -\frac{\sqrt{3}}{2} \]

Therefore, the angle \( \theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \).