Three Dimensional Geometry Question 142
Question: The coordinates of the point p on the line $ \vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda (-\hat{i}+\hat{j}-\hat{k}) $ which is nearest to the origin is
Options:
A) $ ( \frac{2}{3},\frac{4}{3},\frac{2}{3} ) $
B) $ ( -\frac{2}{3},-\frac{4}{3},\frac{2}{3} ) $
C) $ ( \frac{2}{3},\frac{4}{3},-\frac{2}{3} ) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let the point P be (x, y, z), then the vector $ x\hat{i}+y\hat{j}+z\hat{k} $ Will lie on the line. Thus. $ (x-1)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k}=-\lambda \hat{i}+\lambda \hat{j}-\lambda \hat{k} $
$ \Rightarrow x=1-\lambda ,y=1+\lambda $ and $ z=1-\lambda $ Now point P is nearest to the origin
$ \Rightarrow D={{(1-\lambda )}^{2}}+{{(1+\lambda )}^{2}}+{{(1-\lambda )}^{2}} $ Or $ \frac{dD}{d\lambda }=-4(1-\lambda )+2(1+\lambda )=0 $
$ \Rightarrow \lambda =\frac{1}{3} $ Hence, the point is $ ( \frac{2}{3},\frac{4}{3},\frac{2}{3} ) $
BETA
