Vector Algebra Question 94
Question: The image of the point with position vector $ \mathbf{i}+3\mathbf{k} $ in the plane $ \mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=1 $ is
[J & K 2005]
Options:
A) $ \mathbf{i}+2\mathbf{j}+\mathbf{k} $
B) $ \mathbf{i}-2\mathbf{i}+\mathbf{k} $
C) $ -\mathbf{i}-2\mathbf{j}+\mathbf{k} $
D) $ \mathbf{i}+2\mathbf{j}-\mathbf{k} $
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Answer:
Correct Answer: C
Solution:
- Let Q be the image of the point $ P(\mathbf{i}+3\mathbf{k}) $ in the plane $ \mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=1 $ . Then PQ is normal to the plane. Since PQ passes through P and in normal to the given plane, therefore equation of PQ is $ \mathbf{r}=(\mathbf{i}+3\mathbf{k})+\lambda (\mathbf{i}+\mathbf{j}+\mathbf{k}) $
Since, Q lies on the line PQ, so, let the position vector of Q be $ (\mathbf{i}+3\mathbf{k})+\lambda (\mathbf{i}+\mathbf{j}+\mathbf{k}) $
Þ $ (1+\lambda )\mathbf{i}+\lambda \mathbf{j}+(3+\lambda )\mathbf{k} $ . Since R is the mid point of PQ, therefore position vector of R is $ \frac{(1+\lambda )\mathbf{i}+\lambda \mathbf{j}+(3+\lambda )\mathbf{k}+\mathbf{i}+3\mathbf{k}}{2} $ or $ ( \frac{\lambda +2}{2} )\ \mathbf{i}+( \frac{\lambda }{2} )\ \mathbf{j}+( \frac{6+\lambda }{2} )\ \mathbf{k} $ or $ ( \frac{\lambda }{2}+1 )\ \mathbf{i}+( \frac{\lambda }{2} )\ \mathbf{j}+( 3+\frac{\lambda }{2} )\ \mathbf{k} $ Since R lies on the plane $ \mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=1 $ Therefore, $ [ ( \frac{\lambda }{2}+1 )\ \mathbf{i}+( \frac{\lambda }{2} )\ \mathbf{j}+( 3+\frac{\lambda }{2} )\ \mathbf{k} ]\ .\ [ \mathbf{i}+\mathbf{j}+\mathbf{k} ]=1 $ $ [ \frac{\lambda }{2}+1+\frac{\lambda }{2}+3+\frac{\lambda }{2} ]=1 $
Þ $ \lambda =-2 $ So, the position vector of Q is $ (\mathbf{i}+3\mathbf{k})-2(\mathbf{i}+\mathbf{j}+\mathbf{k})=-\mathbf{i}-2\mathbf{j}+\mathbf{k} $ .
BETA
