Express 2 i - j +3 k as the sum of a vector parallel and a vector…

Question

Express $2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ as the sum of a vector parallel and a vector perpendicular to $2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}.$

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Answer

Let $\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\vec{\text{a}}+\vec{\text{b}}\dots(1)$
such that $\vec{\text{a}}$ is a vector parallel to vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$ and $\vec{\text{b}}$ is a vector perpendicular to the vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big).$
since, $\vec{\text{a}}$ is parallel to $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$
$\vec{\text{a}}=\lambda\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$
$\vec{\text{a}}=2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}\dots(2)$
Put value of $\vec{\text{a}}$ in equation (1),
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\big(2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}\big)+\vec{\text{b}}$
$\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}-2\lambda\hat{\text{j}}-4\lambda\hat{\text{j}}+2\lambda\hat{\text{k}}$
$\vec{\text{b}}=(2-2\lambda)\hat{\text{i}}+(-1-4\lambda)\hat{\text{j}}+(3+2\lambda)\hat{\text{k}}$
$\vec{\text{b}}$ is a vector perpendicuar to the vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big),$ then
$\vec{\text{b}}.\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)=0$
$\big[(2-2\lambda)\hat{\text{i}}+(-1-4\lambda)\hat{\text{j}}+(3+2\lambda)\hat{\text{k}}\big]\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)=0$
$(2-2\lambda)(2)+(-1-4\lambda)(4)+(3+2\lambda)(-2)=0$
$4-4\lambda-4-16\lambda-6-4\lambda=0$
$-6-24\lambda=0$
$-24\lambda=6$
$\lambda=-\frac{1}{4}$
Put $\lambda$ in equation (2),
$\vec{\text{a}}=2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}$
$=2\Big(-\frac{1}{4}\Big)\hat{\text{i}}+4\Big(-\frac{1}{4}\Big)\hat{\text{j}}-2\Big(-\frac{1}{4}\Big)\hat{\text{k}}$
$\vec{\text{a}}=-\frac{1}2{}\hat{\text{i}}-\hat{\text{j}}+\frac{1}2{}\hat{\text{k}}$
Put the value of $\vec{\text{a}}$ in equation (1),
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\Big(-\frac{1}{2}\hat{\text{i}}-\hat{\text{j}}+\frac{1}{2}\hat{\text{k}}\Big)+\vec{\text{b}}$
$\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}+\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}-\frac{1}{2}\hat{\text{k}}$
$=\frac{4\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}+\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}{2}$
$=\frac{5\hat{\text{i}}+5\hat{\text{k}}}{2}$
$\vec{\text{b}}=\frac{5}{2}\big(\hat{\text{i}}+\hat{\text{k}}\big)$
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\Big(-\frac{1}{2}\hat{\text{i}}-\hat{\text{j}}+\frac{1}{2}\hat{\text{k}}\Big)+\frac{5}{2}\big(\hat{\text{i}}+\hat{\text{k}}\big)$

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