Vector Algebra Question 221
Question: The centre of the circle given by $ \mathbf{r}.(\mathbf{i}+2\mathbf{j}+2\mathbf{k})=15 $ and $ |\mathbf{r}-(\mathbf{j}+2\mathbf{k})|=4 $ is
Options:
A) (0, 1, 2)
B) (1, 3, 4)
C) (?1, 3, 4)
D) None of these
Show Answer
Answer:
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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