Key Concepts and Formulas

1. Magnitude of a vector $\vec{v} = x\widehat i + y\widehat j + z\widehat k$: $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
2. Dot Product: $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$. Also, $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$.
3. Cross Product Magnitude: $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$.

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Step-by-Step Solution

We are given vectors $\vec{a} = 2\widehat i + \widehat j - 2\widehat k$ and $\vec{b} = \widehat i + \widehat j$. We need to find $|\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c |$ given the conditions: (i) $\vec{a} \cdot \vec{c} = |\vec{c}|$ (ii) $|\vec{c} - \vec{a}| = 2\sqrt{2}$ (iii) The angle between $(\vec{a} \times \vec{b})$ and $\vec{c}$ is $\frac{\pi}{6}$.

Step 1: Calculate the magnitude of vector $\vec{a}$. This is needed for condition (i) and (ii). $|\vec{a}| = \sqrt{(2)^2 + (1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Step 2: Use condition (i) to find the relationship between $\vec{a}$ and $\vec{c}$. We know $\vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta_{ac}$, where $\theta_{ac}$ is the angle between $\vec{a}$ and $\vec{c}$. Given $\vec{a} \cdot \vec{c} = |\vec{c}|$, we have: $|\vec{a}| |\vec{c}| \cos \theta_{ac} = |\vec{c}|$ Since $|\vec{c}| \neq 0$ (otherwise condition (ii) would imply $|\vec{a}| = 2\sqrt{2}$, which is false), we can divide by $|\vec{c}|$: $|\vec{a}| \cos \theta_{ac} = 1$ Substituting $|\vec{a}| = 3$: $3 \cos \theta_{ac} = 1 \implies \cos \theta_{ac} = \frac{1}{3}$

Step 3: Use condition (ii) to find the magnitude of $\vec{c}$. We are given $|\vec{c} - \vec{a}| = 2\sqrt{2}$. Squaring both sides: $|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2 = 8$ Using the property $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$: $(\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = 8$ Expanding the dot product: $\vec{c} \cdot \vec{c} - 2(\vec{a} \cdot \vec{c}) + \vec{a} \cdot \vec{a} = 8$ $|\vec{c}|^2 - 2(\vec{a} \cdot \vec{c}) + |\vec{a}|^2 = 8$ Substitute $|\vec{a}| = 3$ and $\vec{a} \cdot \vec{c} = |\vec{c}|$ from condition (i): $|\vec{c}|^2 - 2|\vec{c}| + 3^2 = 8$ $|\vec{c}|^2 - 2|\vec{c}| + 9 = 8$ $|\vec{c}|^2 - 2|\vec{c}| + 1 = 0$ This is a perfect square: $(|\vec{c}| - 1)^2 = 0$ Therefore, $|\vec{c}| = 1$.

Step 4: Calculate the cross product $\vec{a} \times \vec{b}$ and its magnitude. Let $\vec{P} = \vec{a} \times \vec{b}$. $\vec{P} = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}$ $\vec{P} = \widehat i((1)(0) - (-2)(1)) - \widehat j((2)(0) - (-2)(1)) + \widehat k((2)(1) - (1)(1))$ $\vec{P} = \widehat i(0 + 2) - \widehat j(0 + 2) + \widehat k(2 - 1)$ $\vec{P} = 2\widehat i - 2\widehat j + \widehat k$ Now, calculate its magnitude: $|\vec{P}| = |\vec{a} \times \vec{b}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

Step 5: Calculate the magnitude of the final cross product using condition (iii). We need to find $|\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c | = |\vec{P} \times \vec{c}|$. We are given that the angle between $\vec{P} = (\vec{a} \times \vec{b})$ and $\vec{c}$ is $\theta_{P,c} = \frac{\pi}{6}$. Using the formula for the magnitude of a cross product: $|\vec{P} \times \vec{c}| = |\vec{P}| |\vec{c}| \sin \theta_{P,c}$ Substitute the values we found: $|\vec{P}| = 3$, $|\vec{c}| = 1$, and $\theta_{P,c} = \frac{\pi}{6}$. $|\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c | = (3)(1) \sin\left(\frac{\pi}{6}\right)$ Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$: $|\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c | = 3 \times 1 \times \frac{1}{2} = \frac{3}{2}$

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Common Mistakes & Tips

• Ensure correct application of the dot product and cross product magnitude formulas, especially the angles involved ($\cos \theta$ for dot product, $\sin \theta$ for cross product).
• Be meticulous with algebraic manipulations when expanding vector dot products, particularly when dealing with magnitudes and differences.
• Double-check the calculation of the cross product components, as sign errors are common.

Summary We systematically used the given conditions to find the magnitudes of the involved vectors and the relevant angles. First, we calculated $|\vec{a}|$. Then, condition (i) helped us relate $\vec{a} \cdot \vec{c}$ to $|\vec{c}|$ and $\cos \theta_{ac}$. Condition (ii) was crucial in forming a quadratic equation for $|\vec{c}|$, yielding $|\vec{c}| = 1$. We then computed the cross product $\vec{a} \times \vec{b}$ and found its magnitude to be 3. Finally, using the magnitude of this cross product, the magnitude of $\vec{c}$, and the given angle of $\frac{\pi}{6}$, we calculated the magnitude of the required cross product.

The final answer is $\boxed{\frac{3}{2}}$ which corresponds to option (D).