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

The Correct Option is C

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.