Question
If $\vec{\alpha}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\beta}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}},$ then express $\vec{\beta}$ in the form of $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2,$ where $\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$.

Answer

Given that $\vec{\alpha}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\beta}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
Also,
$\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2,$
$\Rightarrow\vec{\beta}_2=\vec{\beta}+\vec{\beta}_1\dots(1)$
Since $\vec{\beta}_1$ is parallel to $\vec{\alpha},$
$\vec{\beta}_1=\text{t}\vec{\alpha}$
$\Rightarrow\vec{\beta}_1=\text{t}\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)=3\text{t}\hat{\text{i}}+4\text{t}\hat{\text{j}}+5\text{t}\hat{\text{k}}\dots(2)$
Substituting the values of $\vec{\beta}_1$ and $\vec{\alpha}$ in (1),we get
$\vec{\beta}_2=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}-\big(3\text{t}\hat{\text{i}}+4\text{t}\hat{\text{j}}+5\text{t}\hat{\text{k}}\big)\\=(2-3\text{t})\hat{\text{i}}+(1-4\text{t})\hat{\text{j}}+(-4-5\text{t})\hat{\text{k}}\dots(3)$Since $\vec{\beta}_2$ is perpendicular to $\vec{\alpha},$
$\vec{\beta}_2.\vec{\alpha}=0$
$\Rightarrow\Big[(2-3\text{t})\hat{\text{i}}+(1-4\text{t})\hat{\text{j}}+(-4-5\text{t})\hat{\text{k}}\Big].\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)=0$
$\Rightarrow3(2-3\text{t})+4(1-4\text{t})+5(-4-5\text{t})=0$
$\Rightarrow6-9\text{t}+4-16\text{t}-20-25\text{t}=0$
$\Rightarrow-50\text{t}=10$
$\Rightarrow\text{t}=\frac{-1}{5}$
From (2) and (3), we get
$\vec{\beta}_1=\frac{-1}{5}\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)$
$\vec{\beta}_2=\frac{13}{5}\hat{\text{i}}+\frac{9}{5}\hat{\text{j}}-3\hat{\text{k}}=\frac{1}{5}\big(13\hat{\text{i}}+9\hat{\text{j}}-15\hat{\text{k}}\big)$

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