If bar{a}=hat{i}+hat{j} and bar{b}=2 hat{i}-h | vedclass

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(C) Given equations are:$(i)$ $\vec{r} 	imes \vec{a} = \vec{b} 	imes \vec{a}$(ii) $\vec{r} 	imes \vec{b} = \vec{a} 	imes \vec{b}$From (ii),we know

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$(3,1,-1)$

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(C) Given equations are:$(i)$ $\vec{r} 	imes \vec{a} = \vec{b} 	imes \vec{a}$(ii) $\vec{r} 	imes \vec{b} = \vec{a} 	imes \vec{b}$From (ii),we know that $\vec{a} 	imes \vec{b} = -(\vec{b} 	imes \vec{a})$,so $\vec{r} 	imes \vec{b} = -(\vec{b} 	imes \vec{a})$.Adding $(i)$ and (ii):$(\vec{r} 	imes \vec{a}) + (\vec{r} 	imes \vec{b}) = (\vec{b} 	imes \vec{a}) - (\vec{b} 	imes \vec{a}) = \vec{0}$$\vec{r} 	imes (\vec{a} + \vec{b}) = \vec{0}$This implies $\vec{r}$ is parallel to $(\vec{a} + \vec{b})$.Given $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = 2\hat{i} - \hat{k}$,then $\vec{a} + \vec{b} = (1+2)\hat{i} + 1\hat{j} - 1\hat{k} = 3\hat{i} + \hat{j} - \hat{k}$.Thus,$\vec{r} = \lambda(3\hat{i} + \hat{j} - \hat{k})$.For the intersection point,we test the options. For $\lambda = 1$,$\vec{r} = 3\hat{i} + \hat{j} - \hat{k}$,which corresponds to the point $(3, 1, -1)$.Checking this in equation $(i)$: $(3\hat{i} + \hat{j} - \hat{k}) 	imes (\hat{i} + \hat{j}) = 3(\hat{i} 	imes \hat{j}) + 1(\hat{j} 	imes \hat{i}) - 1(\hat{k} 	imes \hat{i}) - 1(\hat{k} 	imes \hat{j}) = 3\hat{k} - \hat{k} - \hat{j} + \hat{i} = \hat{i} - \hat{j} + 2\hat{k}$.And $\vec{b} 	imes \vec{a} = (2\hat{i} - \hat{k}) 	imes (\hat{i} + \hat{j}) = 2(\hat{i} 	imes \hat{j}) - (\hat{k} 	imes \hat{i}) - (\hat{k} 	imes \hat{j}) = 2\hat{k} - \hat{j} + \hat{i} = \hat{i} - \hat{j} + 2\hat{k}$.Since both sides are equal,the point $(3, 1, -1)$ is the intersection point.

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

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