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(D) Let $\overrightarrow{C} = C_1 \hat{i} + C_2 \hat{j} + C_3 \hat{k}$ be a unit vector,so $C_1^2 + C_2^2 + C_3^2 = 1$. Given $\overrightarrow{C} \cdo

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$\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k}$

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(D) Let $\overrightarrow{C} = C_1 \hat{i} + C_2 \hat{j} + C_3 \hat{k}$ be a unit vector,so $C_1^2 + C_2^2 + C_3^2 = 1$. Given $\overrightarrow{C} \cdot (2 \hat{i} + 2 \hat{j} - \hat{k}) = |\overrightarrow{C}| |2 \hat{i} + 2 \hat{j} - \hat{k}| \cos 60^{\circ} = 1 \cdot 3 \cdot \frac{1}{2} = \frac{3}{2}$. So,$2C_1 + 2C_2 - C_3 = \frac{3}{2}$. Also,$\overrightarrow{C} \cdot (\hat{i} - \hat{k}) = |\overrightarrow{C}| |\hat{i} - \hat{k}| \cos 45^{\circ} = 1 \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$. So,$C_1 - C_3 = 1$,which implies $C_3 = C_1 - 1$. Substituting $C_3$ into the first equation: $2C_1 + 2C_2 - (C_1 - 1) = \frac{3}{2} \implies C_1 + 2C_2 = \frac{1}{2} \implies C_2 = \frac{1}{4} - \frac{C_1}{2}$. Using $C_1^2 + C_2^2 + C_3^2 = 1$,we get $C_1^2 + (\frac{1}{4} - \frac{C_1}{2})^2 + (C_1 - 1)^2 = 1$. Solving this quadratic equation yields $C_1 = \frac{1}{2} + \frac{\sqrt{2}}{3}$,$C_2 = -\frac{1}{3\sqrt{2}}$,and $C_3 = \frac{\sqrt{2}}{3} - \frac{1}{2}$. Thus,$\overrightarrow{C} = (\frac{1}{2} + \frac{\sqrt{2}}{3}) \hat{i} - \frac{1}{3\sqrt{2}} \hat{j} + (\frac{\sqrt{2}}{3} - \frac{1}{2}) \hat{k}$. Adding the given vector $(-\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k})$ to $\overrightarrow{C}$ results in $\frac{\sqrt{2}}{3} \hat{i} + 0 \hat{j} - \frac{1}{2} \hat{k} = \frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k}$.

Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{C}$. Then $\overrightarrow{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$ is :

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