QJEEM25Ma74b7: JEE Main 2025 Physics - Class 11 JEEM25Ma74b7 Solution: If the components of along and perpendicular to respectively,…

$\begin{aligned}& \text { let } \\& \overrightarrow{\mathrm{a}}_{11}=\text { component of } \overrightarrow{\mathrm{a}} \text { along } \overrightarrow{\mathrm{b}} \\& \overrightarrow{\mathrm{a}}_1=\text { component of } \overrightarrow{\mathrm{a}} \text { perpendicular to } \overrightarrow{\mathrm{b}} \\& \overrightarrow{\mathrm{a}}_{11}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\& \overrightarrow{\mathrm{a}}_1=\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\& \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_{11}+\overrightarrow{\mathrm{a}}_1 \\& \therefore \overrightarrow{\mathrm{a}}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\& \quad=\frac{44}{11} \hat{\mathrm{i}}+\frac{11}{11} \hat{\mathrm{j}}-\frac{33}{11} \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}} \quad \\& \begin{array}{l}\alpha=4 \qquad \beta=1\qquad \gamma=-3\\\alpha^2+\beta^2+\gamma^2=16+1+9=26\end{array}\end{aligned}$