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,

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

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

Answer 2

To solve this problem, we need to find the magnitude of \(|(\vec{a} \times \vec{b}) \times \vec{c}|\), where \(\vec{a} = 6\hat{i} + \hat{j} - \hat{k}\) and \(\vec{b} = \hat{i} + \hat{j}\), given the conditions involving the vector \(\vec{c}\).

1. Firstly, calculate \(\vec{a} \times \vec{b}\):

\(\hat{i}\)	\(\hat{j}\)	\(\hat{k}\)
6	1	-1
1	1	0

Calculating the determinant:

• For \(\hat{i}:\) \(1(0) - (1)(-1) = 1\)
• For \(\hat{j}:\) \(-(6(0) - 1(-1)) = -1\)
• For \(\hat{k}:\) \(6(1) - 1(1) = 5\)

This gives \(\vec{a} \times \vec{b} = \hat{i} - \hat{j} + 5\hat{k}\).

1. Now, calculate \(| \vec{a} \times \vec{b} |\):

\(| \vec{a} \times \vec{b} | = \sqrt{1^2 + (-1)^2 + 5^2} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3}\)

1. Use the given condition \(\vec{a} \cdot \vec{c} = 6|\vec{c}|\):

\(\vec{a} \cdot \vec{c} = 6c_i + c_j - c_k = 6|\vec{c}| \Rightarrow 6c_i + c_j - c_k = 6\sqrt{c_i^2 + c_j^2 + c_k^2}\)

1. Using the condition \(| \vec{c} - \vec{a} | = 2\sqrt{2}\):

\(\sqrt{(c_i - 6)^2 + (c_j - 1)^2 + (c_k + 1)^2} = 2\sqrt{2}\)

1. From the angle condition, \(\vec{a} \times \vec{b} \cdot \vec{c} = (3\sqrt{3})(6) \cos(60^\circ)\):

\(\vec{a} \times \vec{b} \cdot \vec{c} = 9\sqrt{3}\)

1. The required quantity is:

\(|(\vec{a} \times \vec{b}) \times \vec{c}| = \sqrt{(|\vec{a} \times \vec{b}|^2 |\vec{c}|^2 - (\vec{a} \times \vec{b} \cdot \vec{c})^2)}\)\(= \frac{9}{2}(6 + \sqrt{6})\)

Thus, the answer is \(\frac{9}{2}(6 + \sqrt{6})\).