Question

Let $\vec{a} = 6\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. If $\vec{c}$ is a vector such that \[ |\vec{c}| \geq 6, \quad \vec{a} \cdot \vec{c} = 6 |\vec{c}|, \quad |\vec{c} - \vec{a}| = 2\sqrt{2} \] and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^\circ$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to:

• A. $\frac{9}{2}(6 - \sqrt{6})$
• B. $\frac{3}{2}\sqrt{3}$
• C. $\frac{3}{2}\sqrt{6}$
• D. $\frac{9}{2}(6 + \sqrt{6})$

Answer 1

The Correct Option is D

\[ |(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \cdot \frac{\sqrt{3}}{2} \]

Given:

\[ |\vec{c} - \vec{a}| = 2\sqrt{2} \]

Using the formula for magnitude:

\[ |\vec{c}|^2 + |\vec{a}|^2 - 2 \cdot \vec{a} \cdot \vec{c} = 8 \]

\[ |\vec{c}|^2 + 38 - 12|\vec{c}| = 8 \]

\[ |\vec{c}|^2 - 12|\vec{c}| + 30 = 0 \]

Solving this quadratic equation:

\[ |\vec{c}| = \frac{12 \pm \sqrt{144 - 120}}{2} \]

\[ |\vec{c}| = \frac{12 \pm 2\sqrt{6}}{2} \]

\[ |\vec{c}| = 6 + \sqrt{6} \]

Now, calculating \( \vec{a} \times \vec{b} \):

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix} \]

\[ = -\hat{i} + 7\hat{j} + 5\hat{k} \]

\[ |\vec{a} \times \vec{b}| = \sqrt{27} \]

Thus,

\[ |(\vec{a} \times \vec{b}) \cdot \vec{c}| = \sqrt{27}(6 + \sqrt{6}) \cdot \frac{\sqrt{3}}{2} \]

\[ = \frac{9}{2}(6 + \sqrt{6}) \]