Mathematics - Vectors Question with Solution | TestHub

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