Question

Decompose the vector $6\hat{i}-3\hat{j}-6\hat{k}$ into vectors which are parallel and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}.$

Answer 1

$$\begin{gathered} \text{Let}\overrightarrow{a}\text{=6}\hat{i}-3\hat{j}-6\hat{k}\text{and}\overrightarrow{b}\text{=}\hat{i}+\hat{j}+\hat{k} \\ \mathrm{and}\overrightarrow{x}\text{and}\overrightarrow{y}\text{be such that} \\ \overrightarrow{a}=\overrightarrow{x}+\overrightarrow{y} \\ \Rightarrow \overrightarrow{y}=\overrightarrow{a}-\overrightarrow{x}...\left(1\right) \\ \text{Since}\overrightarrow{x}\text{is parallel to}\overrightarrow{b}\text{,} \\ \overrightarrow{x}=t\overrightarrow{b} \\ \Rightarrow \overrightarrow{x}=t\left(\hat{i}+\hat{j}+\hat{k}\right)=t\hat{i}+t\hat{j}+t\hat{k}\text{...(2)} \\ \text{Substituting the values of}\overrightarrow{x}\text{and}\overrightarrow{a}\text{in (1), we get} \\ \overrightarrow{y}=6\hat{i}-3\hat{j}-6\hat{k}-\left(t\hat{i}+t\hat{j}+t\hat{k}\right)=\left(6-t\right)\hat{i}+\left(-3-t\right)\hat{j}+\left(-6-t\right)\hat{k}...\left(3\right) \\ \text{Since}\overrightarrow{y}\text{is perpendicular to}\overrightarrow{b}\text{,} \\ \overrightarrow{y}.\overrightarrow{b}=0 \\ \Rightarrow \left[\left(6-t\right)\hat{i}+\left(-3-t\right)\hat{j}+\left(-6-t\right)\hat{k}\right].\left(\hat{i}+\hat{j}+\hat{k}\right)=0 \\ \Rightarrow 1\left(6-t\right)+1\left(-3-t\right)+1\left(-6-t\right)=0 \\ \Rightarrow -3-3t=0 \\ \Rightarrow t=-1 \\ \text{From (2) and (3), we get} \\ \overrightarrow{x}=-\hat{i}-\hat{j}-\hat{k} \\ \overrightarrow{y}=7\hat{i}-2\hat{j}-5\hat{k} \\ \text{So,} \\ \overrightarrow{a}=\overrightarrow{x}+\overrightarrow{y}=\left(-\hat{i}-\hat{j}-\hat{k}\right)+\left(7\hat{i}-2\hat{j}-5\hat{k}\right) \end{gathered}$$