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Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+\hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=((\overrightarrow{\mathrm{a}} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\overrightarrow{\mathrm{a}}$ on $\overrightarrow{\mathrm{b}}$ is :
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$\begin{aligned} & \vec{a} \times(\hat{i}+\hat{j})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 0\end{array}\right| \\ & =\hat{i}-\hat{j}+\hat{k} \\ & (\vec{a} \times(\hat{i} \times \hat{j})) \times \hat{i}=\hat{k}+\hat{j} \\ & ((\vec{a} \times(\hat{i} \times \hat{j})) \times i) \times \hat{i}=\hat{j}-\hat{k} \\ & \operatorname{projection} \text { of } \vec{a} \text { on } \hat{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\ & =\frac{1+1}{\sqrt{2}}=\sqrt{2}\end{aligned}$
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