The unit vector vec{r} which satisfies vec{r} | vedclass

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(A) Given the equation $\vec{r} 	imes \vec{b} = \vec{r} 	imes \vec{c}$.This can be rewritten as $\vec{r} 	imes \vec{b} - \vec{r} 	imes \vec{c} = \

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$\pm \left( \frac{2\hat{i} - 2\hat{j} + \hat{k}}{3} \right)$

Vedclass · Answer

(A) Given the equation $\vec{r} 	imes \vec{b} = \vec{r} 	imes \vec{c}$.This can be rewritten as $\vec{r} 	imes \vec{b} - \vec{r} 	imes \vec{c} = \vec{0}$,which implies $\vec{r} 	imes (\vec{b} - \vec{c}) = \vec{0}$.Calculate $\vec{b} - \vec{c} = (\hat{i} + 2\hat{j} + \hat{k}) - (3\hat{i} + 2\hat{k}) = -2\hat{i} + 2\hat{j} - \hat{k}$.Since $\vec{r} 	imes (\vec{b} - \vec{c}) = \vec{0}$,the vector $\vec{r}$ must be parallel to $(\vec{b} - \vec{c})$.Thus,$\vec{r} = \lambda (-2\hat{i} + 2\hat{j} - \hat{k})$.For $\vec{r}$ to be a unit vector,we have $|\vec{r}| = 1$,so $|\lambda| \sqrt{(-2)^2 + 2^2 + (-1)^2} = 1$.$|\lambda| \sqrt{4 + 4 + 1} = 1 \Rightarrow |\lambda| \sqrt{9} = 1 \Rightarrow 3|\lambda| = 1 \Rightarrow \lambda = \pm \frac{1}{3}$.Therefore,$\hat{r} = \pm \frac{1}{3} (-2\hat{i} + 2\hat{j} - \hat{k}) = \pm \left( \frac{2\hat{i} - 2\hat{j} + \hat{k}}{3} ight)$.

The unit vector $\vec{r}$ which satisfies $\vec{r} \times \vec{b} = \vec{r} \times \vec{c}$,where $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 3\hat{i} + 2\hat{k}$,is:

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