QJEEM24Ma7d23: JEE Main 2024 Physics - Class 11 JEEM24Ma7d23 Solution: Let . Then the square of the projection of on is:

$\begin{aligned}& \vec{a}=2 \hat{i}+\hat{j}-\hat{k} \\& \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i} \\& \vec{a} \times(\hat{i}+\hat{j})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\2 & 1 & -1 \\1 & 1 & 0\end{array}\right| \\&=\hat{i}(1)-\hat{j}(1)+\hat{k}(2-1) \\&=\hat{i}-\hat{j}+\hat{k}\end{aligned}$

$\begin{aligned}& (\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & -1 & 1 \\1 & 0 & 0\end{array}\right| \\& \quad=\hat{i}(0)-\hat{j}(-1)+\hat{k}(1) \\& \quad=\hat{j}+\hat{k}\end{aligned}$

$$\left((\overrightarrow{a}\times (\hat{i}+\hat{j})\times \hat{i}\right)\times \hat{i}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 1\\ 1& 0& 0\end{array}\right|$$

$\begin{aligned}& =\hat{i}(0)-\hat{j}(-1)+\hat{k}(-1) \\\vec{b}= & \hat{j}-\hat{k}\end{aligned}$

Projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

$=\frac{2}{\sqrt{2}}=\sqrt{2}$

Square of projection $=2$