Question
Decompose the vector $6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}$ into vectors which are parallal and perpendicular to the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
✓
Answer
Let $\vec{\text{a}}=6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
and $\vec{\text{x}}$ 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 parallel to $\vec{\text{b}},$
$\vec{\text{x}}=\text{t}\vec{\text{b}}$
$\Rightarrow\vec{\text{x}}=\text{t}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}+\text{t}\hat{\text{k}}\dots(2)$
Substituting the values of $\vec{\text{x}}$ and $\vec{\text{a}}$ in (1), we get
$\vec{\text{y}}=6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}-\big(\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}+\text{t}\hat{\text{k}}\big)=(6-\text{t})\hat{\text{i}}+(-3-\text{t})\hat{\text{j}}+(-6-\text{t})\hat{\text{k}}\dots(3)$
Since $\vec{\text{y}}$ is perpendicular to $\vec{\text{b}},$
$\vec{\text{y}}.\vec{\text{b}}=0$
$\Rightarrow\big[(6-\text{t})\hat{\text{i}}+(-3-\text{t})\hat{\text{j}}+(-6-\text{t})\hat{\text{k}}\big].\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=0$
$\Rightarrow1(6-\text{t})+1(-3-\text{t})+1(-6-\text{t})=0$
$\Rightarrow-3-3\text{t}=0$
$\Rightarrow\text{t}=-1$
From (2) and (3), we get
$\vec{\text{x}}=-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{y}}=7\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$
So,
$\vec{\text{a}}=\vec{\text{x}}+\vec{\text{y}}=\big(-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\big(7\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}\big)$
and $\vec{\text{x}}$ 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 parallel to $\vec{\text{b}},$
$\vec{\text{x}}=\text{t}\vec{\text{b}}$
$\Rightarrow\vec{\text{x}}=\text{t}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}+\text{t}\hat{\text{k}}\dots(2)$
Substituting the values of $\vec{\text{x}}$ and $\vec{\text{a}}$ in (1), we get
$\vec{\text{y}}=6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}-\big(\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}+\text{t}\hat{\text{k}}\big)=(6-\text{t})\hat{\text{i}}+(-3-\text{t})\hat{\text{j}}+(-6-\text{t})\hat{\text{k}}\dots(3)$
Since $\vec{\text{y}}$ is perpendicular to $\vec{\text{b}},$
$\vec{\text{y}}.\vec{\text{b}}=0$
$\Rightarrow\big[(6-\text{t})\hat{\text{i}}+(-3-\text{t})\hat{\text{j}}+(-6-\text{t})\hat{\text{k}}\big].\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=0$
$\Rightarrow1(6-\text{t})+1(-3-\text{t})+1(-6-\text{t})=0$
$\Rightarrow-3-3\text{t}=0$
$\Rightarrow\text{t}=-1$
From (2) and (3), we get
$\vec{\text{x}}=-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{y}}=7\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$
So,
$\vec{\text{a}}=\vec{\text{x}}+\vec{\text{y}}=\big(-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\big(7\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}\big)$
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