If the components of a= i+ j+ k along and perpendicular to b=3i+j-k respectively, are (16)…

Question

If the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3\hat{i}+\hat{j}-\hat{k}$ 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 :

Accepted Answer

$\begin{gathered} {\overset{	o }{a}}_{11}=\overset{	o }{a}\overset{	o }{b} \ {\overset{	o }{a}}_{1}=\overset{	o }{a}\overset{	o }{b} \ {\overset{	o }{a}}_{11}=\frac{16}{11}(3\hat{i}+\hat{j}-\hat{k}) \ {\overset{	o }{a}}_{1}=\frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k}) \ \because \overset{	o }{a}={\overset{	o }{a}}_{11}+{\overset{	o }{a}}_{1} \ 	herefore \overset{	o }{a}=\frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})+\frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k}) \ \, =\frac{44}{11}\hat{i}+\frac{11}{11}\hat{j}-\frac{33}{11}\hat{k} \ \overset{	o }{a}=4\hat{i}+\hat{j}-3\hat{k}\, \ \begin{gathered} \alpha =4\, \beta =1\, \gamma =-3 \ {\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}=16+1+9=26 \end{gathered} \end{gathered}$

Solutions · 1