A) \[\mathbf{r}.(\mathbf{i}+9\mathbf{j}+11\mathbf{k})=0\]
B) \[\mathbf{r}.(\mathbf{i}+9\mathbf{j}+11\mathbf{k})=6\]
C) \[\mathbf{r}.(\mathbf{i}-3\mathbf{j}-13\mathbf{k})=0\]
D) None of these
Correct Answer: A
Solution :
The vector equation of a plane through the line of intersection of the planes \[\mathbf{r}.(\mathbf{i}+3\mathbf{j}-\mathbf{k})=0\] and \[\mathbf{r}.(\mathbf{j}+2\mathbf{k})\]=0 can be written as \[(\mathbf{r}.(\mathbf{i}+3\mathbf{j}-\mathbf{k}))+\lambda (\mathbf{r}.(\mathbf{j}+2\mathbf{k}))=0\] .....(i) This passes through \[2\mathbf{i}+\mathbf{j}-\mathbf{k}\] \ \[(2\mathbf{i}+\mathbf{j}-\mathbf{k}).(\mathbf{i}+3\mathbf{j}-\mathbf{k})+\lambda (2\mathbf{i}+\mathbf{j}-\mathbf{k}).(\mathbf{j}+2\mathbf{k})=0\] or \[(2+3+1)+\lambda (0+1-2)=0\Rightarrow \lambda =6\] Put the value of \[\lambda \] in (i) we get \[\mathbf{r}.(\mathbf{i}+9\mathbf{j}+11\mathbf{k})=0\], which is the required plane.
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