The point of intersection of $\overrightarrow{\mathbf{r}} imes \overrightarrow{\mathbf{a}} = \overrightarrow{\mathbf{b}} imes \overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{r}} imes \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{a}} imes \overrightarrow{\mathbf{b}},$ where $\overrightarrow{\mathbf{a}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}$ and $\overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} - \widehat{\mathbf{k}}$ is

Question

The point of intersection of $\overrightarrow{\mathbf{r}} 	imes \overrightarrow{\mathbf{a}} = \overrightarrow{\mathbf{b}} 	imes \overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{r}} 	imes \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}},$ where $\overrightarrow{\mathbf{a}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}$ and $\overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} - \widehat{\mathbf{k}}$ is

Dr. Abubacker · Accepted Answer

(a)
The point of intersection of $\overrightarrow{\mathbf{r}} 	imes \overrightarrow{\mathbf{a}} = \overrightarrow{\mathbf{b}} 	imes \overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{r}} 	imes \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}}$ is $\overrightarrow{\mathbf{r}} = \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}$
$	herefore\ \overrightarrow{\mathbf{r}} = \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} ight) + (2\widehat{\dot{\mathbf{i}}} - \widehat{\mathbf{k}}) = 3\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$

Explanation

Reviewed by Dr. Abubacker

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