A) \[\mathbf{r}.(2\mathbf{i}+7\mathbf{j}-13\mathbf{k})=1\]
B) \[\mathbf{r}.(2\mathbf{i}-7\mathbf{j}-13\mathbf{k})=1\]
C) \[\mathbf{r}.(2\mathbf{i}+7\mathbf{j}+13\mathbf{k})=0\]
D) None of these
Correct Answer: B
Solution :
The line of intersection of the planes \[\mathbf{r}.(3\mathbf{i}-\mathbf{j}+\mathbf{k})=1\] and \[\mathbf{r}.(\mathbf{i}+4\mathbf{j}-2\mathbf{k})=2\] is common to both the planes. Therefore, it is perpendicular to normals to the two planes i.e., \[{{\mathbf{n}}_{1}}=3\mathbf{i}-\mathbf{j}+\mathbf{k}\] and \[{{\mathbf{n}}_{2}}=\mathbf{i}+4\mathbf{j}-2\mathbf{k}\]. Hence it is parallel to the vector \[{{\mathbf{n}}_{1}}\times {{\mathbf{n}}_{2}}=-2\mathbf{i}+7\mathbf{j}+13\mathbf{k}.\] Thus, we have to find the equation of the plane passing through \[\mathbf{a}=\mathbf{i}+2\mathbf{j}-\mathbf{k}\]and normal to the vector \[\mathbf{n}={{\mathbf{n}}_{1}}\times {{\mathbf{n}}_{2}}\]. The equation of the required plane is \[(\mathbf{r}-\mathbf{a}).\ \mathbf{n}=0\] or \[\mathbf{r}.\,\mathbf{n}=\mathbf{a}.\mathbf{n}\] or \[\mathbf{r}.(-2\mathbf{i}+7\mathbf{j}+13\mathbf{k})\]=\[(\mathbf{i}+2\mathbf{j}-\mathbf{k}).(-2\mathbf{i}+7\mathbf{j}+13\mathbf{k})\] or \[\mathbf{r}.(2\mathbf{i}-7\mathbf{j}-13\mathbf{k})=1\].
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