Let vec{a}=hat{i} and vec{b}=hat{j}. The poin | vedclass

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(A) Given,$\vec{r} 	imes \vec{a} = \vec{b} 	imes \vec{a}$.This implies $(\vec{r} - \vec{b}) 	imes \vec{a} = 0$,which means $\vec{r} - \vec{b}$ is p

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$\vec{r}=\hat{i}+\hat{j}$

Vedclass · Answer

(A) Given,$\vec{r} 	imes \vec{a} = \vec{b} 	imes \vec{a}$.This implies $(\vec{r} - \vec{b}) 	imes \vec{a} = 0$,which means $\vec{r} - \vec{b}$ is parallel to $\vec{a}$.So,the equation of the first line is $\vec{r} = \vec{b} + p\vec{a} = \hat{j} + p\hat{i}$.Similarly,for the second line $\vec{r} 	imes \vec{b} = \vec{a} 	imes \vec{b}$,we have $(\vec{r} - \vec{a}) 	imes \vec{b} = 0$.This means $\vec{r} - \vec{a}$ is parallel to $\vec{b}$.So,the equation of the second line is $\vec{r} = \vec{a} + q\vec{b} = \hat{i} + q\hat{j}$.For the point of intersection,$\hat{j} + p\hat{i} = \hat{i} + q\hat{j}$.Comparing the coefficients of $\hat{i}$ and $\hat{j}$,we get $p = 1$ and $q = 1$.Substituting $p=1$ in the first equation,we get $\vec{r} = \hat{j} + 1(\hat{i}) = \hat{i} + \hat{j}$.

Let $\vec{a}=\hat{i}$ and $\vec{b}=\hat{j}$. 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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