If the components of $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\overrightarrow{\mathrm{b}}=3 \hat{i}+\hat{j}-\hat{k}$ respectively, are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$, then…

Question

If the components of $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\overrightarrow{\mathrm{b}}=3 \hat{i}+\hat{j}-\hat{k}$ respectively, are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$, then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Aicharya · Accepted Answer

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

If the components of $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and…

Solution