Three Dimensional Geometry Question 239
Question: A line with direction cosines proportional to 2,1, 2 meets each of the lines $ x=y+a=z $ and $ x+a=2y=2z $ . The co-ordinates of each of the points of intersection are given by
[AIEEE 2004]
Options:
A) $ (2a,a,,3a),(2a,,a,,a) $
B) $ (3a,,2a,,3a),\ (a,,a,,a) $
C) $ (3a,,2a,,3a),(a,,a,,2a) $
D) $ (3a,,3a,,3a),(a,,a,,a) $
Show Answer
Answer:
Correct Answer: B
Solution:
Let the two lines be AB and CD having equation $ \frac{x}{1}=\frac{y+a}{1}=\frac{z}{1}=\lambda $ and $ \frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}=\mu $ then $ P\equiv (\lambda ,,\lambda -a,\lambda ) $ and $ Q=(2\mu -a,,\mu ,,\mu ) $ So according to question, $ \frac{\lambda -2\mu +a}{2}=\frac{\lambda -a-\mu }{1} $ $ =\frac{\lambda -\mu }{2} $
Þ $ \mu =a $ and $ \lambda =3a $
$ \therefore $ $ P\equiv (3a,,2a,,3a) $ and $ [{{(x-2)}^{2}}+{{(y-3)}^{2}}+{{(z-4)}^{2}}] $ . Trick: Put the options and check it.
BETA
