Let a→=i^+j^ and b→=2i^−k^ then the point of intersection of the lines r→×a→=b→×a→ and r→×b→=a→×b→ is

let r→×a→=b→×a→ ⇒(r→−b→)×a→=0→⇒r→=b→+ta→Similarly, other line r→=a→+kb→ where t and k are scalars.now a→+kb→=b→+ta→⇒ t=1,k=1∴ r→=a→+b→=i^+j^+2i^−k^=3i^+j^−k^i.e. (3, 1, -1)

Let a→=i^+j^ and b→=2i^−k^ then the point of intersection of the lines r→×a→=b→×a→ and r→×b→=a→×b→ is

let r→×a→=b→×a→ ⇒(r→−b→)×a→=0→⇒r→=b→+ta→Similarly, other line r→=a→+kb→ where t and k are scalars.now a→+kb→=b→+ta→⇒ t=1,k=1∴ r→=a→+b→=i^+j^+2i^−k^=3i^+j^−k^i.e. (3, 1, -1)

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Let $\vec{a} = \hat{i} + \hat{j} and \vec{b} = 2 \hat{i} - \hat{k}$ then the point of intersection of the lines $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is

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(3,- 1, 1)

(3, 1, -1)

(-3,1, 1)

(-3,- 1, -1)

answer is B.

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Detailed Solution

$\begin{array}{l} let \vec{r} \times \vec{a} = \vec{b} \times \vec{a} \\ \Rightarrow (\vec{r} - \vec{b}) \times \vec{a} = \vec{0} \Rightarrow \vec{r} = \vec{b} + t \vec{a} \end{array}$