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

r×a=b×a=-(a×b) r×b=a×b r×a+r×b=0 ⇒ r×(a+b)=0 r || a+b ⇒ r=λ(i+3j-k) ⇒ |r|=λ1+9+1 ⇒ r|r|=i+3j-k11

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

r×a=b×a=-(a×b) r×b=a×b r×a+r×b=0 ⇒ r×(a+b)=0 r || a+b ⇒ r=λ(i+3j-k) ⇒ |r|=λ1+9+1 ⇒ r|r|=i+3j-k11

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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

$r \times a = b \times a = - (a \times b) r \times b = a \times b r \times a + r \times b = 0 \Rightarrow r \times (a + b) = 0 r | | a + b \Rightarrow r = λ (i + 3 j - k) \Rightarrow | r | = λ \sqrt{1 + 9 + 1} \Rightarrow \frac{r}{| r |} = \frac{i + 3 j - k}{\sqrt{11}}$