If $\bar{a}=\hat{i}+\hat{j}, \bar{b}=2 \hat{i}-\hat{k}$, then point of intersection of the lines $\bar{r} imes \bar{a}=\bar{b} imes \bar{a}$ and $\bar{r} imes \bar{b}=\bar{a} imes \bar{b}$ is

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If $\bar{a}=\hat{i}+\hat{j}, \bar{b}=2 \hat{i}-\hat{k}$, then point of intersection of the lines $\bar{r} 	imes \bar{a}=\bar{b} 	imes \bar{a}$ and $\bar{r} 	imes \bar{b}=\bar{a} 	imes \bar{b}$ is

Aicharya · Accepted Answer

$\vec{r} 	imes \vec{a}=\vec{b} 	imes \vec{a}$ and $\vec{r} 	imes \vec{b}=\vec{a} 	imes \vec{b}$ $\begin{aligned} & \Rightarrow \vec{r} 	imes \vec{a}=-\vec{r} 	imes \vec{b} \quad 	ext { [from (i) and (ii)] } \ & \Rightarrow \vec{r} 	imes(\vec{a}+\vec{b})=\overrightarrow{0} \ & \Rightarrow \vec{r} \| \vec{a}+\vec{b} \ & \Rightarrow \vec{r}=\lambda(\vec{a}+\vec{b})=\lambda(3 \hat{i}+\hat{j}-\hat{k})\end{aligned}$ Taking $\lambda=1, \vec{r}=3 \hat{i}+\hat{j}-\widehat{k} \equiv(3,1,-1)$

If $\bar{a}=\hat{i}+\hat{j}, \bar{b}=2 \hat{i}-\hat{k}$, then point of intersection of the lines $\bar{r}…

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