If overline{a}=hat{i}+hat{j},overline{b}=2hat | vedclass

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(D) Given $\overline{r} 	imes \overline{a} = \overline{b} 	imes \overline{a} \implies (\overline{r} - \overline{b}) 	imes \overline{a} = \overline{

Vedclass · Accepted Answer

$\frac{\hat{i}+3\hat{j}-\hat{k}}{\sqrt{11}}$

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(D) Given $\overline{r} 	imes \overline{a} = \overline{b} 	imes \overline{a} \implies (\overline{r} - \overline{b}) 	imes \overline{a} = \overline{0}$. This implies $(\overline{r} - \overline{b}) = k_1 \overline{a}$ for some scalar $k_1$.Similarly,$\overline{r} 	imes \overline{b} = \overline{a} 	imes \overline{b} \implies (\overline{r} - \overline{a}) 	imes \overline{b} = \overline{0}$. This implies $(\overline{r} - \overline{a}) = k_2 \overline{b}$ for some scalar $k_2$.From the first equation,$\overline{r} = \overline{b} + k_1 \overline{a} = (2\hat{j} - \hat{k}) + k_1(\hat{i} + \hat{j}) = k_1\hat{i} + (2+k_1)\hat{j} - \hat{k}$.From the second equation,$\overline{r} = \overline{a} + k_2 \overline{b} = (\hat{i} + \hat{j}) + k_2(2\hat{j} - \hat{k}) = \hat{i} + (1+2k_2)\hat{j} - k_2\hat{k}$.Comparing the components:$k_1 = 1$$2+k_1 = 1+2k_2 \implies 2+1 = 1+2k_2 \implies 2k_2 = 2 \implies k_2 = 1$$-1 = -k_2 \implies k_2 = 1$.Substituting $k_1=1$ into $\overline{r} = k_1\hat{i} + (2+k_1)\hat{j} - \hat{k}$,we get $\overline{r} = \hat{i} + 3\hat{j} - \hat{k}$.The magnitude $|\overline{r}| = \sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{1+9+1} = \sqrt{11}$.Therefore,$\frac{\overline{r}}{|\overline{r}|} = \frac{\hat{i} + 3\hat{j} - \hat{k}}{\sqrt{11}}$.

If $\overline{a}=\hat{i}+\hat{j}$,$\overline{b}=2\hat{j}-\hat{k}$ and $\overline{r} \times \overline{a}=\overline{b} \times \overline{a}$,$\overline{r} \times \overline{b}=\overline{a} \times \overline{b}$,then the value of $\frac{\overline{r}}{|\overline{r}|}$ is

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