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

Answer: (A)

$\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

Answer 2

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 $\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.