Three Dimensional Geometry Question 223
Question: In $ \Delta ABC $ the mid-point of the sides AB, BC and CA are respectively (l, 0, 0), (0, m, 0) and (0, 0, n). Then, $ \frac{AB^{2}+BC^{2}+CA^{2}}{l^{2}+m^{2}+n^{2}} $ is equal to
Options:
A) 2
B) 4
C) 8
D) 16
Show Answer
Answer:
Correct Answer: C
Solution:
[c] From the figure $ x_1+x_2=21,y_1+y_2=0,z_1+z_2=0 $ $ x_2+x_3=0,y_2+y_3=2m,z_1+z_3=0 $ and $ x_1+x_3=0,y_1+y_3=0,z_1+z_3=2n $ On solving, we get $ x_1=l,x_2=l,x_3=-1 $ $ y_1=-m,y_2=m,y_3=m $ and $ z_1=n,z_2=-n,z_3=n $
$ \therefore $ Coordinates are $ A(l,-m,n),,B(l,m,-n) $ and $ C(-l,m,n) $
$ \therefore \frac{AB^{2}+BC^{2}+CA^{2}}{l^{2}+m^{2}+n^{2}} $ $ =\frac{(4m^{2}+4n^{2})+(4l^{2}+4n^{2})+(4l^{2}+4m^{2})}{l^{2}+m^{2}+n^{2}}=8 $
BETA
