If vec{a} = hat{i} + hat{j} and vec{b} = hat{ | vedclass

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(A) Given equations are $\vec{r} 	imes \vec{a} = \vec{b} 	imes \vec{a} \implies (\vec{r} - \vec{b}) 	imes \vec{a} = 0$ and $\vec{r} 	imes \vec{b}

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$(2, 1, -1)$

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(A) Given equations are $\vec{r} 	imes \vec{a} = \vec{b} 	imes \vec{a} \implies (\vec{r} - \vec{b}) 	imes \vec{a} = 0$ and $\vec{r} 	imes \vec{b} = \vec{a} 	imes \vec{b} \implies (\vec{r} - \vec{a}) 	imes \vec{b} = 0$.This implies $\vec{r} - \vec{b} = t\vec{a}$ and $\vec{r} - \vec{a} = s\vec{b}$ for some scalars $t, s$.Thus, $\vec{r} = \vec{b} + t\vec{a} = (\hat{i} - \hat{k}) + t(\hat{i} + \hat{j}) = (1+t)\hat{i} + t\hat{j} - \hat{k}$.Also, $\vec{r} = \vec{a} + s\vec{b} = (\hat{i} + \hat{j}) + s(\hat{i} - \hat{k}) = (1+s)\hat{i} + \hat{j} - s\hat{k}$.Equating the components: $1+t = 1+s \implies t=s$, $t=1$, and $-1 = -s \implies s=1$.Substituting $t=1$ into $\vec{r} = (1+t)\hat{i} + t\hat{j} - \hat{k}$, we get $\vec{r} = 2\hat{i} + \hat{j} - \hat{k}$.The point is $(2, 1, -1)$.

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

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