Three Dimensional Geometry Question 408
Question: If lines $ x=y=z $ and $ x=\frac{y}{2}=\frac{z}{3} $ and third line passing through (1, 1, 1) form a triangle of area $ \sqrt{6} $ units, then the point of intersection of third line with the second line will be
Options:
A) $ (1,2,3) $
B) $ (2,4,6) $
C) $ ( \frac{4}{3},\frac{8}{3},\frac{12}{3} ) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let any point on the second line be $ (\lambda ,2\lambda ,3\lambda ) $ $ \cos \theta =\frac{6}{\sqrt{42}},\sin \theta =\frac{\sqrt{6}}{\sqrt{42}} $ $ {\Delta_{OAB}}=\frac{1}{2}(OA)OBsin\theta $ $ =\frac{1}{2}\sqrt{3}\lambda \sqrt{14}\times \frac{\sqrt{6}}{\sqrt{42}}=\sqrt{6}or\lambda =2 $ So, B is (2, 4, 6)
BETA
