Question

For any vector $\overrightarrow{r}$ prove that $\overrightarrow{r}=$ $(\overrightarrow{r}.\overrightarrow{i})\overrightarrow{r}+(\overrightarrow{r}.\overrightarrow{j})\overrightarrow{j}+(\overrightarrow{r}.\overrightarrow{k})\overrightarrow{k}$

Answer 1

To prove : $\overrightarrow{r}=(\overrightarrow{r}.\overrightarrow{c})\overrightarrow{r}+(\overrightarrow{r}.\overrightarrow{j})\overrightarrow{j}+(\overrightarrow{r}.\overrightarrow{k})\overrightarrow{k}$

Solving R.H.S $\overrightarrow{r}=(x\hat{i}+y\hat{j}+z\hat{k})$

So,$(\overrightarrow{r},\overrightarrow{i})=\left[\right(x\hat{i}+y\hat{j}+z\hat{k}).\hat{i}]\hat{i}+\left[\right(x\hat{i}+y\hat{j}+z\hat{k}).\hat{j}]\hat{j}+\left[\right(x\hat{i}+y\hat{j}+z\hat{k}).\hat{k}]\hat{k}$

$=[x\hat{i},\hat{i}+y\hat{j}.\hat{i}+z\hat{k}.\hat{i}]\hat{i}+[x\hat{i}.\hat{j}+y\hat{j}.\widehat{j+2\hat{k}.\hat{j}}]\hat{i}+[x\hat{i}.\hat{k}+y\hat{j}.\hat{k}+2\hat{k}.\hat{k}]\hat{k}$

$=[\because \hat{i}.\hat{i}=1=\hat{j}.\hat{j}=\hat{k}.\hat{k}$ & $\hat{i}.\hat{j}=\hat{j},\hat{k}=\hat{k}.\hat{i}=0]$

$=x\hat{i}+y\hat{j}+z\hat{k}$

$=r=R.H.S$

$\Rightarrow L.H.S=R.H.S$

Hence proved.