Question

If $\overrightarrow{a}$ is any vector, then show that $\overrightarrow{a}$=$(\overrightarrow{a}.\hat{i})$$\hat{i}+(\overrightarrow{a}.\hat{j})\hat{j}+(\overrightarrow{a}.\hat{k})\hat{k}.$

Answer 1

Let $\overrightarrow{a}$ be the unit vector.

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

Then,

$(a.\hat{i}).\hat{i}+(b.\hat{j}).\hat{j}+(c.\hat{k}).\hat{k}$

$\Rightarrow [(x\hat{i}+y\hat{j}+z\hat{k}).\hat{i}].\hat{i}+[(x\hat{i}+y\hat{j}+z\hat{k}).\hat{j}].\hat{j}+[(x\hat{i}+y\hat{j}+z\hat{k}).\hat{k}].\hat{k}$

Now, using

$\hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1$

$\hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}=0$

We get,

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

Hence, proved.