Question
$\text{if} \overrightarrow{\text{r}} = x\hat{\text{i}} + y\hat{\text{j}} + z\hat{\text{k}}, \text{find} \overrightarrow(\text{r} \times \hat{\text{i}}). (\overrightarrow{\text{r}} \times \text{j}) + xy$

Answer

$\overrightarrow{\text{r}}\times \overrightarrow{\text{i}} = \bigg(\text{x}\hat{\text{i}} + \text{y}\hat{\text{j}} + \text{z}\hat{\text{k}}\bigg)\text{x}\hat{\text{i}} = -\text{y}\hat{\text{k}} + \text{z}\hat{\text{j}}$
$\overrightarrow{\text{r}}\times\overrightarrow{\text{j}} = \bigg(\text{x}\hat{\text{i}} + \text{y}\hat{\text{j}} + \text{z}\hat{\text{k}}\bigg)\hat{\text{j}} = \text{x}\hat{\text{k}} - \text{z}\hat{\text{i}}$
$\bigg(\overrightarrow{\text{r}}\times\hat{\text{i}}\bigg), \bigg(\overrightarrow{\text{r}}\times\overrightarrow{\text{j}}\bigg) = \bigg(\text{o}\hat{\text{i}} + \text{z}\hat{\text{j}} - \text{y}\hat{\text{k}}\bigg).\bigg(\text{- z}\hat{\text{i}} + \text{o}\hat{\text{j}} + \text{x}\hat{\text{k}}\bigg)= -\text{xy}$
$\bigg(\overrightarrow{\text{r}}\times\hat{\text{i}}\bigg). \bigg(\overrightarrow{\text{r}}\times\overrightarrow{\text{j}}\bigg) + \text{xy} = \text{-xy + xy = 0}$

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