Question

Express the vector $\overrightarrow{a}=5\hat{i}+5\hat{k}$ as sum of two vector such that one is parallel to the $\overrightarrow{b}=3\hat{i}+\hat{k}$ and other is perpendicular to $\overrightarrow{b}$

Answer 1

Let the vector parallel to $\overrightarrow{b}=3\hat{i}+\hat{k}$ is $n(3\hat{i}+\hat{k})$ and the vector perpendicular to $\overrightarrow{b}$ be $x\hat{i}+y\hat{j}+z\hat{k}$

Given that adding these two vectors will give $\overrightarrow{a}=5\hat{i}+5\hat{k}$

$\therefore n(3\hat{i}+\hat{k})+x\hat{i}+y\hat{j}+z\hat{k}=5\hat{i}+5\hat{k}$

$\Rightarrow 3n\hat{i}+n\hat{k}+x\hat{i}+y\hat{j}+z\hat{k}=5\hat{i}+5\hat{k}$

Therefore,

$x+3n=5.....\left(1\right)$

$y=0$

$z+n=5.....\left(2\right)$

Since vector $x\hat{i}+y\hat{j}+z\hat{k}$ is perpendicular to $\overrightarrow{b}$, their dot product must be zero.

$\therefore (3\hat{i}+\hat{k}).(x\hat{i}+y\hat{j}+z\hat{k})=0$

$\Rightarrow 3x+z=0$

$\Rightarrow z=-3x.....\left(3\right)$

Substituting the value of z in ${eq}^n\left(2\right)$, we have

$-3x+n=5.....\left(4\right)$

From ${eq}^n\left(1\right)\&\left(4\right)$, we get

$n=2$

$x=-1$

On substituting the value of x in ${eq}^n\left(2\right)$, we get

$z=-3\times (-1)=3$

Hence the vector parallel to $\overrightarrow{b}$ will be $2(3\hat{i}+\hat{k})\Rightarrow 6\hat{i}+2\hat{k}$ and the vector perpendicular to $\overrightarrow{b}$ will be $-1\hat{i}+0\hat{j}+3\hat{k}\Rightarrow -\hat{i}+3\hat{k}$.