If vec{a} is a vector such that vec{a} times | vedclass

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(A) Let $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}$.Given $\vec{a} 	imes \hat{i} = \hat{j} + \hat{k}$.Calculating the cross product: $\vec{a} 	ime

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$\vec{r}=(t+1) \hat{i}+(1-t) \hat{j}+(t+1) \hat{k}$

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(A) Let $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}$.Given $\vec{a} 	imes \hat{i} = \hat{j} + \hat{k}$.Calculating the cross product: $\vec{a} 	imes \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) 	imes \hat{i} = -y \hat{k} + z \hat{j} = z \hat{j} - y \hat{k}$.Comparing with $\hat{j} + \hat{k}$, we get $z = 1$ and $y = -1$.Given $\vec{a} \cdot \hat{i} = 1$, so $(x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{i} = x = 1$.Thus, $\vec{a} = \hat{i} - \hat{j} + \hat{k}$.The equation of the line passing through point $\vec{p} = \hat{i} + \hat{j} + \hat{k}$ and parallel to $\vec{a}$ is $\vec{r} = \vec{p} + t \vec{a}$.$\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + t(\hat{i} - \hat{j} + \hat{k}) = (1+t) \hat{i} + (1-t) \hat{j} + (1+t) \hat{k}$.

If $\vec{a}$ is a vector such that $\vec{a} \times \hat{i}=\hat{j}+\hat{k}$ and $\vec{a} \cdot \hat{i}=1$, then the equation of the line passing through the point $\hat{i}+\hat{j}+\hat{k}$ and parallel to $\vec{a}$ is

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