Question

Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}| = 2, |\vec{b}| = 3$ and $\vec{a} = \vec{b} \times \vec{c}$. If $\alpha \in [0, \frac{\pi}{3}]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c}| - |\vec{a}|^2$ is equal to:

• A. 110
• B. 105
• C. 124
• D. 121

Answer 1

Answer: (C)

124

Answer 2

Given:

$|\vec{a}| = 2, \, |\vec{b}| = 3, \, \vec{a} = \vec{b} \times \vec{c}, \, \text{and } \alpha \text{ is the angle between } \vec{b} \text{ and } \vec{c}.$

From the cross product property:

$|\vec{a}| = |\vec{b}||\vec{c}| \sin \alpha,$

we have:

$2 = 3 \cdot |\vec{c}| \cdot \sin \alpha \implies |\vec{c}| = \frac{2}{3 \sin \alpha}.$

Next, we calculate $|\vec{c} - \vec{a}|^2$:

$|\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2 (\vec{c} \cdot \vec{a}).$

Using $|\vec{c}| = \frac{2}{3 \sin \alpha}$ and $|\vec{a}| = 2$, we find:

$|\vec{c}|^2 = \left(\frac{2}{3 \sin \alpha}\right)^2 = \frac{4}{9 \sin^2 \alpha}, \quad |\vec{a}|^2 = 4.$

For $\vec{c} \cdot \vec{a}$, since $\vec{a} \perp \vec{b}$ (as $\vec{a} = \vec{b} \times \vec{c}$), we have:

$\vec{c} \cdot \vec{a} = 0.$

Thus:

$|\vec{c} - \vec{a}|^2 = \frac{4}{9 \sin^2 \alpha} + 4.$

The expression to minimize is:

$27|\vec{c} - \vec{a}|^2 = 27 \left(\frac{4}{9 \sin^2 \alpha} + 4 \right).$

Simplify:

$27|\vec{c} - \vec{a}|^2 = 12 \csc^2 \alpha + 108.$

To minimize, note that $\csc^2 \alpha$ is minimized when $\sin \alpha$ is maximized. The maximum value of $\sin \alpha$ in the interval $\alpha \in \left[0, \frac{\pi}{3}\right]$ is $\sin \alpha = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:

$\csc^2 \alpha = \frac{1}{\sin^2 \alpha} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{4}{3}.$

Substitute:

$27|\vec{c} - \vec{a}|^2 = 12 \cdot \frac{4}{3} + 108 = 16 + 108 = 124.$

Therefore, the minimum value of $27|\vec{c} - \vec{a}|^2$ is 124.