Vector Algebra Question 133
Question: The position vector of the point where the line $ \mathbf{r}=\mathbf{i}-\mathbf{j}+\mathbf{k}+t(\mathbf{i}+\mathbf{j}+\mathbf{k}) $ meets the plane $ \mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=5 $ is
[Kerala (Engg.) 2005]
Options:
A) $ 5\mathbf{i}+\mathbf{j}-\mathbf{k} $
B) $ 5\mathbf{i}+3\mathbf{j}-3\mathbf{k} $
C) $ 2\mathbf{i}+\mathbf{j}+2\mathbf{k} $
D) $ 5\mathbf{i}+\mathbf{j}+\mathbf{k} $
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Answer:
Correct Answer: B
Solution:
- The point of the given line is $ (1+t,-1+t,1-t) $ Equation of plane is, $ x+y+z=5 $ The point of the given line satisfies the equation of plane
$ \therefore $ $ (1+t)(-1+t)+(1-t)=5 $
Þ $ 1+t=5 $
Þ $ t=4 $ \ Points are $ (5,3,,-3) $ Hence, position vector of point is, $ 5\mathbf{i}+3\mathbf{j}-3\mathbf{k} $ .
BETA
