Key Concepts and Formulas

• Magnitude of a Cross Product: For any two vectors $\vec{X}$ and $\vec{Y}$, the magnitude of their cross product is given by $|\vec{X} \times \vec{Y}| = |\vec{X}| |\vec{Y}| \sin \theta$, where $\theta$ is the angle between $\vec{X}$ and $\vec{Y}$.
• Vector Triple Product: The magnitude of the vector triple product of vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ can be expressed as $|(\vec{A} \times \vec{B}) \times \vec{C}| = |\vec{A} \times \vec{B}| |\vec{C}| \sin \phi$, where $\phi$ is the angle between the vector $\vec{A} \times \vec{B}$ and the vector $\vec{C}$.
• Dot Product and Magnitude: For any vector $\vec{V}$, $|\vec{V}|^2 = \vec{V} \cdot \vec{V}$. Also, if $\vec{V} \cdot \vec{W} = k|\vec{W}|$, and $\vec{W} \neq \vec{0}$, then the projection of $\vec{V}$ onto $\vec{W}$ is $k$.

Step-by-Step Solution

Step 1: Calculate the cross product $\vec{a} \times \vec{b}$. We are given $\vec{a} = 6\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. The cross product $\vec{a} \times \vec{b}$ is calculated as: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix}$ $\vec{a} \times \vec{b} = \hat{i}(1 \cdot 0 - (-1) \cdot 1) - \hat{j}(6 \cdot 0 - (-1) \cdot 1) + \hat{k}(6 \cdot 1 - 1 \cdot 1)$ $\vec{a} \times \vec{b} = \hat{i}(0 + 1) - \hat{j}(0 + 1) + \hat{k}(6 - 1)$ $\vec{a} \times \vec{b} = \hat{i} - \hat{j} + 5\hat{k}$

Step 2: Calculate the magnitude of $\vec{a} \times \vec{b}$. Let $\vec{X} = \vec{a} \times \vec{b} = \hat{i} - \hat{j} + 5\hat{k}$. The magnitude of $\vec{X}$ is: $|\vec{X}| = |\vec{a} \times \vec{b}| = \sqrt{1^2 + (-1)^2 + 5^2}$ $|\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3}$

Step 3: Use the given information about $\vec{a} \cdot \vec{c}$ and $|\vec{c} - \vec{a}|$. We are given $\vec{a} \cdot \vec{c} = 6|\vec{c}|$. This implies that the projection of $\vec{a}$ onto $\vec{c}$ has a magnitude of $6$. We are also given $|\vec{c} - \vec{a}| = 2\sqrt{2}$. Squaring both sides, we get: $|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2$ $(\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = 8$ $\vec{c} \cdot \vec{c} - 2(\vec{a} \cdot \vec{c}) + \vec{a} \cdot \vec{a} = 8$ $|\vec{c}|^2 - 2(\vec{a} \cdot \vec{c}) + |\vec{a}|^2 = 8$ First, let's find $|\vec{a}|^2$: $|\vec{a}|^2 = 6^2 + 1^2 + (-1)^2 = 36 + 1 + 1 = 38$ Now substitute the values into the equation: $|\vec{c}|^2 - 2(6|\vec{c}|) + 38 = 8$ $|\vec{c}|^2 - 12|\vec{c}| + 30 = 0$ This is a quadratic equation in terms of $|\vec{c}|$. Let $x = |\vec{c}|$. $x^2 - 12x + 30 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $|\vec{c}| = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(30)}}{2(1)}$ $|\vec{c}| = \frac{12 \pm \sqrt{144 - 120}}{2}$ $|\vec{c}| = \frac{12 \pm \sqrt{24}}{2}$ $|\vec{c}| = \frac{12 \pm 2\sqrt{6}}{2}$ $|\vec{c}| = 6 \pm \sqrt{6}$ We are given that $|\vec{c}| \geq 6$. Both $6 + \sqrt{6}$ and $6 - \sqrt{6}$ satisfy this condition if $\sqrt{6} \ge 0$. However, the equation $|\vec{a} \cdot \vec{c}| = 6|\vec{c}|$ from the dot product definition $\vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta$ implies $|\vec{a}| |\cos \theta| = 6$. Since $|\vec{a}| = \sqrt{38}$, $|\cos \theta| = \frac{6}{\sqrt{38}} < 1$. The relation $\vec{a} \cdot \vec{c} = 6|\vec{c}|$ means $|\vec{a}||\vec{c}| \cos \theta = 6|\vec{c}|$, so $|\vec{a}|\cos\theta = 6$. Thus $\cos\theta = \frac{6}{\sqrt{38}}$. The relation $|\vec{c}| = 6 \pm \sqrt{6}$ comes from the quadratic equation. Let's re-examine the condition $|\vec{a} \cdot \vec{c}| = 6|\vec{c}|$. This means $|\vec{a}||\vec{c}||\cos\theta|=6|\vec{c}|$. So $|\vec{a}||\cos\theta|=6$, which is $\sqrt{38}|\cos\theta|=6$. This is consistent. The quadratic equation arises from $|\vec{c}-\vec{a}|^2=8$. $|\vec{c}|^2 - 2 \vec{a} \cdot \vec{c} + |\vec{a}|^2 = 8$. $|\vec{c}|^2 - 2 (6|\vec{c}|) + 38 = 8$. $|\vec{c}|^2 - 12|\vec{c}| + 30 = 0$. The solutions are $|\vec{c}| = 6+\sqrt{6}$ and $|\vec{c}| = 6-\sqrt{6}$. Since $|\vec{c}| \ge 6$, we must have $|\vec{c}| = 6+\sqrt{6}$.

Step 4: Use the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$. Let $\vec{Y} = \vec{a} \times \vec{b}$. We are given that the angle between $\vec{Y}$ and $\vec{c}$ is $60^{\circ}$. This means $\phi = 60^{\circ}$.

Step 5: Calculate the magnitude of the vector triple product $|(\vec{a} \times \vec{b}) \times \vec{c}|$. We use the formula for the magnitude of a cross product: $|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \sin(60^{\circ})$ We have $|\vec{a} \times \vec{b}| = 3\sqrt{3}$ from Step 2 and $|\vec{c}| = 6 + \sqrt{6}$ from Step 3. And $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$. Substituting these values: $|(\vec{a} \times \vec{b}) \times \vec{c}| = (3\sqrt{3}) \left(6 + \sqrt{6}\right) \left(\frac{\sqrt{3}}{2}\right)$ $|(\vec{a} \times \vec{b}) \times \vec{c}| = \frac{3 \cdot 3}{2} \left(6 + \sqrt{6}\right)$ $|(\vec{a} \times \vec{b}) \times \vec{c}| = \frac{9}{2} \left(6 + \sqrt{6}\right)$

Common Mistakes & Tips

• Algebraic Errors: Be careful with algebraic manipulations, especially when squaring vectors and solving quadratic equations.
• Misinterpreting Dot Product Conditions: The condition $\vec{a} \cdot \vec{c} = 6|\vec{c}|$ is crucial. It relates the magnitudes and the angle between $\vec{a}$ and $\vec{c}$.
• Quadratic Formula Application: Ensure the quadratic formula is applied correctly when solving for $|\vec{c}|$.

Summary

The problem requires us to find the magnitude of the vector triple product $|(\vec{a} \times \vec{b}) \times \vec{c}|$. We first computed the cross product $\vec{a} \times \vec{b}$ and its magnitude. Then, using the given conditions $|\vec{c}| \geq 6$, $\vec{a} \cdot \vec{c}=6|\vec{c}|$, and $|\vec{c}-\vec{a}|=2 \sqrt{2}$, we derived a quadratic equation for $|\vec{c}|$ and found its value to be $6+\sqrt{6}$. Finally, using the formula for the magnitude of the cross product and the given angle between $\vec{a} \times \vec{b}$ and $\vec{c}$, we calculated the desired magnitude.

The final answer is $\boxed{\frac{9}{2}(6+\sqrt{6})}$.