question_answer
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]
A)
\[(2a,\,\,a,\,3a),(2a,\,a,\,a)\]
B)
\[(3a,\,2a,\,3a),\ (a,\,a,\,a)\]
C)
\[(3a,\,2a,\,3a),(a,\,a,\,2a)\]
D)
\[(3a,\,3a,\,3a),(a,\,a,\,a)\]
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.
