A) (0, 1, 2)
B) (1, 3, 4)
C) (?1, 3, 4)
D) None of these
Correct Answer: B
Solution :
The equation of a line through the centre \[\mathbf{j}+2\mathbf{k}\] and normal to the given plane is \[\mathbf{r}=\mathbf{j}+2\mathbf{k}+\lambda (\mathbf{i}+2\mathbf{j}+2\mathbf{k})\] .....(i) This meets the plane at a point for which we must have \[((\mathbf{j}+2\mathbf{k})+\lambda (\mathbf{i}+2\mathbf{j}+2\mathbf{k})).(\mathbf{i}+2\mathbf{j}+2\mathbf{k})=15\] Þ \[6+\lambda (9)=15\Rightarrow \lambda =1\]. Putting \[\lambda =1\] in (i), we obtain the position vector of the centre as \[\mathbf{i}+3\mathbf{j}+4\mathbf{k}\]. Hence, the coordinates of the centre of the circle are (1, 3, 4).
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