Question

For any vector $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ , prove that the value of $|\overrightarrow{a}\times \overrightarrow{i}{|}^2+|\overrightarrow{a}\times \overrightarrow{j}{|}^2+|\overrightarrow{a}\times \overrightarrow{k}{|}^2=|2\overrightarrow{a}{|}^2$.

Answer 1

Consider the problem

Let $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$

thus,

$|\overrightarrow{a}|=a_1^2+a_2^2+a_3^2$ ----- from (i)

Now,

$\overrightarrow{a}\times \hat{i}=(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\times \hat{i}$

$=-a_2\hat{k}+a_3\hat{j}$

$|\overrightarrow{a}\times \hat{i}{|}^2={a_2}^2+{a_3}^2$

$\overrightarrow{a}\times \hat{j}=(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\times \hat{j}=a_1\hat{k}-a_3\hat{i}$

$|\overrightarrow{a}\times \hat{j}{|}^2={a_1}^2+{a_3}^2$

$\overrightarrow{a}\times \hat{k}=(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\times \hat{k}=-a_{1}j+a_{2}i$

$|\overrightarrow{a}\times \hat{k}{|}^2={a_1}^2+{a_2}^2$

Now

$|\overrightarrow{a}\times \overrightarrow{i}{|}^2+|\overrightarrow{a}\times \overrightarrow{j}{|}^2+|\overrightarrow{a}\times \overrightarrow{k}{|}^2$

$=a_2^2+a_3^2+a_1^2+a_3^2+a_1^2+a_2^2$

$=2\left[a_1^2+a_2^2+a_3^2\right]$

$=2|\overrightarrow{a}{|}^2$ from (i)

Hence prove .