Let $$\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

<p>$$\begin{aligned} & \vec{a}=2 \hat{i}+\hat{j}-\hat{k} \\ & \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i} \\ & \vec{a} \times(\hat{i}+\hat{j})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 0 \end{array}\right| \\ &=\hat{i}(1)-\hat{j}(1)+\hat{k}(2-1) \\ &=\hat{i}-\hat{j}+\hat{k} \end{aligned}$$</p> <p>$$\begin{aligned} & (\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 0 & 0 \end{array}\right| \\ & \quad=\hat{i}(0)-\hat{j}(-1)+\hat{k}(1) \\ & \quad=\hat{j}+\hat{k} \end{aligned}$$</p> <p>$$\left( {(\overrightarrow a \times (\widehat i + \widehat j) \times \widehat i} \right) \times \widehat i = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right|$$</p> <p>$$\begin{aligned} & =\hat{i}(0)-\hat{j}(-1)+\hat{k}(-1) \\ \vec{b}= & \hat{j}-\hat{k} \end{aligned}$$</p> <p>Projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$</p> <p>$=\frac{2}{\sqrt{2}}=\sqrt{2}$</p> <p>Square of projection $=2$</p>