A) \[\mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=0\]
B) \[\mathbf{r}.(\mathbf{i}-\mathbf{j}-\mathbf{k})=0\]
C) \[\mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=3\]
D) None of these
Correct Answer: B
Solution :
Given two lines \[\mathbf{r}=(\mathbf{i}+\mathbf{j})+\lambda (\mathbf{i}+2\mathbf{j}-\mathbf{k})\] and \[\mathbf{r}=(\mathbf{i}+\mathbf{j})+\mu (-\mathbf{i}+\mathbf{j}-2\mathbf{k})\]pass through \[\mathbf{a}=\mathbf{i}+\mathbf{j}\] and are parallel to the vectors \[\mathbf{b}=\mathbf{i}+2\mathbf{j}-\mathbf{k}\]and \[\mathbf{c}=-\mathbf{i}+\mathbf{j}-2\mathbf{k}\] respectively. Therefore the plane containing them passes through \[\mathbf{a}=\mathbf{i}+\mathbf{j}\] and is perpendicular to \[\mathbf{n}=\mathbf{b}\times \mathbf{c}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})\times (-\mathbf{i}+\mathbf{j}-2\mathbf{k})=-3\mathbf{i}+3\mathbf{j}+3\mathbf{k}\]. Hence, the equation of the plane is \[(\mathbf{r}-\mathbf{a}).\mathbf{n}=0\Rightarrow \mathbf{r}.\mathbf{n}=\mathbf{a}.\mathbf{n}\Rightarrow \mathbf{r}.(\mathbf{i}-\mathbf{j}-\mathbf{k})=0\].
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