Question

Express $2\overline{i}+3\overline{j}+\overline{k}$ as sum of two vectors out of which one vector is perpendicular to $2\overline{i}-4\overline{j}+\overline{k}$ and another is parallel to $2\overline{i}-4\overline{j}+\overline{k}$.

Answer 1

Let vector parallel to $2\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}$ is $d(2\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=2d\overrightarrow{i}-4d\overrightarrow{j}+d\overrightarrow{k}$

Let vector perpendicular to $2\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}$ is$a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}$ such that $2a+4b+c=0$

Now $2\overrightarrow{i}+3\overrightarrow{j}+\overrightarrow{k}=2d\overrightarrow{i}-4d\overrightarrow{j}+d\overrightarrow{k}+a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}$

Equating $\overrightarrow{i},\overrightarrow{j},\overrightarrow{k}$ terms

$2=2d+a$

$2-2d=a$ (1)

$3=-4d+b$

$3+4d=b$ (2)

$1=d+c$

$1-d=c$ (3)

Also $2a-4b+c=0$ (4)

Substituting (1),(2) and (3) in (4)

$2(2-2d)-4(3+4d)+1-d=0$

$4-4d-12-16d+1-d=0$

$-7-21d=0$

$d=\frac{-1}{3}$ (5)

Substituting (5) in (1),(2) and (3)

$2-2\times \frac{-1}{3}=a$

$a=\frac{8}{3}$

$3+4\times \frac{-1}{3}=b$

$b=\frac{5}{3}$

$1-d=c$

$1-\frac{-1}{3}=c$

$c=\frac{4}{3}$

Hence the vector which is perpendicular is $\frac{8}{3}\overrightarrow{i}+\frac{5}{3}\overrightarrow{j}+\frac{4}{3}\overrightarrow{k}$

The vector which is parallel is $\frac{-2}{3}\overrightarrow{i}+\frac{4}{3}\overrightarrow{j}+\frac{-1}{3}\overrightarrow{k}$