Three Dimensional Geometry Question 60
Question: The centre of the circle given by $ \vec{r}\cdot (\hat{i}+2\hat{j}+2\hat{k})=15 $ and $ | \vec{r}-(\hat{j}+2\hat{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:
[b] The equation of the line through the centre $ \hat{j}+2\hat{k} $ and normal to the given plane is $ \vec{r}=\hat{j}+2\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k}) $ …(i) This meets the plane for which $ [\hat{j}+2\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})]\cdot (\hat{i}+2\hat{j}+2\hat{k})=15 $ Or $ 6+9\lambda =15 $ or $ \lambda =1 $ Putting in (i) we get $ \vec{r}=\hat{j}+2\hat{k}+(\hat{i}+2\hat{j}+2\hat{k}) $ $ =\hat{i}+3\hat{j}+4\hat{k} $ Hence, centre is $ (1,3,4) $ .
BETA
