Question

Express the vector $\overrightarrow{a}=5\hat{i}-2\hat{j}+5\hat{k}$ as the sum of two vectors such that one is parallel to the vector $\overrightarrow{b}=3\hat{i}+\hat{k}$ and other is perpendicular to $\overrightarrow{b}$.

Answer 1

$$\begin{gathered} \text{Given that}\overrightarrow{a}\text{=}5\hat{i}-2\hat{j}+5\hat{k}\text{and}\overrightarrow{b}\text{=3}\hat{i}\text{+}\hat{k} \\ \text{Let}\overrightarrow{x}\text{and}\overrightarrow{y}\text{be such that} \\ \overrightarrow{a}=\overrightarrow{x}+\overrightarrow{y} \\ \Rightarrow \overrightarrow{y}=\overrightarrow{a}-\overrightarrow{x}...\left(1\right) \\ \text{Since}\overrightarrow{x}\text{is parallel to}\overrightarrow{b}\text{,} \\ \Rightarrow \overrightarrow{x}=t\overrightarrow{b}\left(t\mathrm{is}\mathrm{constant}\right) \\ \Rightarrow \overrightarrow{x}=t\left(3\hat{i}\text{+}\hat{k}\right)=3t\hat{i}\text{+t}\hat{k} \\ \text{Substituting the values of}\overrightarrow{x}\text{and}\overrightarrow{a}\text{in (1), we get} \\ \overrightarrow{y}=5\hat{i}-2\hat{j}+5\hat{k}-\left(3t\hat{i}\text{+t}\hat{k}\right)=\left(5-3t\right)\hat{i}-2\hat{j}+\left(5-t\right)\hat{k}...\left(2\right) \\ \text{Since}\overrightarrow{y}\text{is perpendicular to}\overrightarrow{b}\text{,} \\ \overrightarrow{y}.\overrightarrow{b}=0 \\ \Rightarrow \left[\left(5-3t\right)\hat{i}-2\hat{j}+\left(5-t\right)\hat{k}\right].\left(3\hat{i}\text{+}\hat{k}\right)=0 \\ \Rightarrow 3\left(5-3t\right)+0+\left(5-t\right)=0 \\ \Rightarrow 15-9t+5-t=0 \\ \Rightarrow 20-10t=0 \\ \Rightarrow t=2 \\ \text{From (1) and (2), we get} \\ \overrightarrow{x}=6\hat{i}\text{+2}\hat{k} \\ \overrightarrow{y}=-\hat{i}-2\hat{j}+3\hat{k} \end{gathered}$$