Let a¯=i¯+j¯,b¯=2j¯−k¯. If r¯×a¯=b¯×a¯,r¯×b¯=a¯×b¯then r¯r¯=

r¯×a¯=−a¯×b¯r¯×b¯=a¯×b¯r¯×a¯+r¯+b¯=0¯r¯×(a¯+b¯)=0r¯=t(a¯+b¯)

Let a¯=i¯+j¯,b¯=2j¯−k¯. If r¯×a¯=b¯×a¯,r¯×b¯=a¯×b¯then r¯r¯=

r¯×a¯=−a¯×b¯r¯×b¯=a¯×b¯r¯×a¯+r¯+b¯=0¯r¯×(a¯+b¯)=0r¯=t(a¯+b¯)

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Let $\bar{a} = \bar{i} + \bar{j}, \bar{b} = 2 \bar{j} - \bar{k} .$ If $\bar{r} \times \bar{a} = \bar{b} \times \bar{a}, \bar{r} \times \bar{b} = \bar{a} \times \bar{b}$ then $\frac{\bar{r}}{|\bar{r}|} =$

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$\frac{1}{\sqrt{11}} (\bar{i} + 3 \bar{j} - \bar{k})$

$\frac{1}{\sqrt{11}} (\bar{i} - 3 \bar{j} + \bar{k})$

$\frac{1}{\sqrt{3}} (\bar{i} - \bar{j} + \bar{k})$

$\frac{1}{\sqrt{3}} (\bar{i} + \bar{j} - \bar{k})$

answer is A.

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Detailed Solution

$\begin{aligned} \bar{r} \times \bar{a} = - \bar{a} \times \bar{b} \\ \bar{r} \times \bar{b} = \bar{a} \times \bar{b} \\ \bar{r} \times \bar{a} + \bar{r} + \bar{b} = \bar{0} \\ \bar{r} \times (\bar{a} + \bar{b}) = 0 \end{aligned}$
$\begin{aligned} \bar{r} = t (\bar{a} + \bar{b}) \end{aligned}$