Question
Express the vector $\vec{\text{a}}=5\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}$ as the sum of two vectors such that one is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{k}}$ and other is perpendicular to $\vec{\text{b}}.$

Answer

Given that $\vec{\text{a}}=5\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{k}}$
Let $\vec{\text{x}}$ and $\vec{\text{y}} $ be such that
$\vec{\text{a}} =\vec{\text{x}} +\vec{\text{y}} $
$\Rightarrow\vec{\text{y}} =\vec{\text{a}} -\vec{\text{x}} \dots(1)$
Since $\vec{\text{x}}$ is parallei to $\vec{\text{b}},$
$\Rightarrow\vec{\text{x}}=\text{t}\vec{\text{b}}$ (t is constant)
$\Rightarrow\vec{\text{x}}=\text{t}\big(3\hat{\text{i}}+\vec{\text{k}}\big)=3\text{t}\hat{\text{i}}+\text{t}\hat{\text{k}}$
Substituting the values of $\vec{\text{x}}$ and $\vec{\text{a}}$ in (1), we get
$\vec{\text{y}}=5\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}-\big(3\text{t}\hat{\text{i}}+\text{t}\hat{\text{k}}\big)=(5-3\text{t})\hat{\text{i}}-2\hat{\text{j}}+(5-\text{t})\hat{\text{k}}\dots(2)$
Since $\vec{\text{y}} $ is perpendicular to $\vec{\text{b}},$
$\vec{\text{y}}.\vec{\text{b}}=0$
$\Rightarrow\big[(5-3\text{t})\hat{\text{i}}-2\hat{\text{j}}+(5-\text{t})\hat{\text{k}}\big].\big(3\hat{\text{i}}+\hat{\text{k}}\big)=0$
$\Rightarrow3(5-3\text{t})+0+(5-\text{t})=0$
$\Rightarrow15-9\text{t}+5-\text{t}=0$
$\Rightarrow20-10\text{t}=0$
$\Rightarrow\text{t}=2$
From (1) and (2),we get
$\vec{\text{x}}=6\hat{\text{i}}+2\hat{\text{k}}$
$\vec{\text{y}}=-\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$

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