Key Concepts and Formulas

• Vector Magnitude: The magnitude of a vector $\vec{v} = v_x \widehat{i} + v_y \widehat{j} + v_z \widehat{k}$ is $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.
• Cross Product: The cross product of two vectors $\vec{u}$ and $\vec{v}$ is given by $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$, where $\theta$ is the angle between them.
• Dot Product and Magnitude: The square of the magnitude of a vector difference is $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$.

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Step-by-Step Solution

Step 1: Calculate the magnitude of $\overrightarrow a$.

• Why: The magnitude of $\overrightarrow a$ is needed to use the formula for the magnitude of a vector difference in a later step.
• Given $\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$.
• $|\overrightarrow a| = \sqrt{(2)^2 + (1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Step 2: Calculate the cross product $\overrightarrow a \times \overrightarrow b$.

• Why: The vector $\overrightarrow a \times \overrightarrow b$ is explicitly used in the given conditions, so we need to compute it.
• Given $\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$ and $\overrightarrow b = \widehat i + \widehat j$.
• $\overrightarrow a \times \overrightarrow b = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}$
• $= \widehat i ((1)(0) - (-2)(1)) - \widehat j ((2)(0) - (-2)(1)) + \widehat k ((2)(1) - (1)(1))$
• $= \widehat i (0 + 2) - \widehat j (0 + 2) + \widehat k (2 - 1)$
• $\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$

Step 3: Calculate the magnitude of $\overrightarrow a \times \overrightarrow b$.

• Why: The magnitude of $\overrightarrow a \times \overrightarrow b$ is required to apply the cross product magnitude formula.
• From Step 2, $\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$.
• $|\overrightarrow a \times \overrightarrow b| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

Step 4: Determine the magnitude of $\overrightarrow c$ using the cross product condition.

• Why: We are given a condition involving the magnitude of a cross product and the angle between the vectors, which allows us to find $|\overrightarrow c|$.
• We are given $|\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c | = 3$ and the angle between $\overrightarrow c$ and $\overrightarrow a \times \overrightarrow b$ is $30^\circ$.
• Let $\vec{p} = \overrightarrow a \times \overrightarrow b$. We have $|\vec{p}| = 3$ (from Step 3) and the angle $\theta = 30^\circ$.
• Using the formula $|\vec{p} \times \overrightarrow c| = |\vec{p}| |\overrightarrow c| \sin \theta$:
• $3 = (3) |\overrightarrow c| \sin 30^\circ$
• $3 = 3 |\overrightarrow c| \left(\frac{1}{2}\right)$
• Dividing by 3: $1 = |\overrightarrow c| \left(\frac{1}{2}\right)$
• Multiplying by 2: $|\overrightarrow c| = 2$

Step 5: Calculate $\overrightarrow a \cdot \overrightarrow c$ using the first given condition.

• Why: The condition $|\overrightarrow c - \overrightarrow a| = 3$ involves the difference between $\overrightarrow c$ and $\overrightarrow a$, and its square can be expanded to include the dot product $\overrightarrow a \cdot \overrightarrow c$.
• Given $|\overrightarrow c - \overrightarrow a| = 3$.
• Squaring both sides: $|\overrightarrow c - \overrightarrow a|^2 = 3^2 = 9$
• Using the formula $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$:
• $|\overrightarrow c|^2 + |\overrightarrow a|^2 - 2(\overrightarrow c \cdot \overrightarrow a) = 9$
• Since $\overrightarrow c \cdot \overrightarrow a = \overrightarrow a \cdot \overrightarrow c$, we have:
• $|\overrightarrow c|^2 + |\overrightarrow a|^2 - 2(\overrightarrow a \cdot \overrightarrow c) = 9$
• Substitute the values $|\overrightarrow c| = 2$ (from Step 4) and $|\overrightarrow a| = 3$ (from Step 1):
• $(2)^2 + (3)^2 - 2(\overrightarrow a \cdot \overrightarrow c) = 9$
• $4 + 9 - 2(\overrightarrow a \cdot \overrightarrow c) = 9$
• $13 - 2(\overrightarrow a \cdot \overrightarrow c) = 9$
• Subtract 13 from both sides:
• $-2(\overrightarrow a \cdot \overrightarrow c) = 9 - 13$
• $-2(\overrightarrow a \cdot \overrightarrow c) = -4$
• Divide by -2:
• $\overrightarrow a \cdot \overrightarrow c = \frac{-4}{-2} = 2$

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Common Mistakes & Tips

• Cross Product Calculation: Be meticulous when calculating the cross product using the determinant method, especially with signs.
• Magnitude of Difference Expansion: Ensure the correct expansion of $|\vec{x} - \vec{y}|^2$ as $|\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$, not just $|\vec{x}|^2 - |\vec{y}|^2$.
• Angle Interpretation: The angle given in the cross product condition is between $\overrightarrow c$ and $\overrightarrow a \times \overrightarrow b$.

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Summary

The problem is solved by systematically calculating intermediate vector quantities like $|\overrightarrow a|$ and $\overrightarrow a \times \overrightarrow b$. The magnitude of $\overrightarrow c$ is derived from the condition involving the magnitude of a cross product and the angle between vectors. Finally, the dot product $\overrightarrow a \cdot \overrightarrow c$ is found by expanding the squared magnitude of the vector difference $|\overrightarrow c - \overrightarrow a|$ and substituting the known magnitudes.

The final answer is $\boxed{2}$.