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 the other is perpendicular to $\overrightarrow{b}$.

Answer 1

$\begin{array}{c}\overrightarrow{a}=5\hat{i}-2\hat{j}+5\hat{k}\\ \\ \overrightarrow{b}=3\hat{i}+\hat{k}\end{array}$

Consider,

$\overrightarrow{a}=\overrightarrow{{\beta }_1}+\overrightarrow{{\beta }_2}$

Where $\overrightarrow{{\beta }_1}$ is parallel to $\overrightarrow{b}$ and $\overrightarrow{{\beta }_2}$ is perpendicular to $\overrightarrow{b}$

Since,

$\overrightarrow{{\beta }_1}$ is parallel to $\overrightarrow{b}$

Then,

$\overrightarrow{{\beta }_1}=\lambda \overrightarrow{b}=\lambda (3\hat{i}+\hat{k})=3\lambda \hat{i}+\lambda \hat{k}$ ------ $\left(1\right)$

Now,

$\overrightarrow{a}=\overrightarrow{{\beta }_1}+\overrightarrow{{\beta }_2}$

$\overrightarrow{{\beta }_2}=\overrightarrow{a}-\overrightarrow{{\beta }_1}$

$\overrightarrow{{\beta }_2}=(5\hat{i}-2\hat{j}+5\hat{k})-(3\lambda \hat{i}+\lambda \hat{k})$

$\overrightarrow{{\beta }_2}=(5-3\lambda )\hat{i}-2\hat{j}+(5-\lambda )\hat{k}$ ----- $\left(2\right)$

Since,

$\overrightarrow{{\beta }_2}$ is perpendicular to $\overrightarrow{b}$

$\overrightarrow{{\beta }_2}.\overrightarrow{b}=0$

Then,

$3(5-3\lambda )+0(-2)+1(5-\lambda )=0$

$\lambda =2$

Therefore,

$\overrightarrow{{\beta }_1}=6\hat{i}+2\hat{k}$ from $\left(1\right)$

$\overrightarrow{{\beta }_2}=-\hat{i}-2\hat{j}+3\hat{k}$ ----- $\left(2\right)$

Hence,

$\overrightarrow{a}=(6\hat{i}+2\hat{k})+(-\hat{i}-2\hat{j}+3\hat{k})$