Three Dimensional Geometry Question 4
Question: If from a point $ P(a,b,c) $ perpendiculars $ PA $ and $ PB $ are drawn to yz and zx planes, then the equation of the plane $ OAB $ is
Options:
A) $ bcx+cay+abz=0 $
B) $ bcx+cay-abz=0 $
C) $ bcx-cay+abz=0 $
D) $ -bcx+cay+abz=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ A,(0,b,c) $ in yz-plane and $ B,(a,0,c) $ in zx-plane. Plane through O is $ px+qy+rz=0. $ It passes through A and B.
$ \therefore 0p+qb+rc=0 $ and $ pa+0q+rc=0 $
$ \Rightarrow \frac{p}{bc}=\frac{q}{ca}=\frac{r}{-ab}=k $
$ \Rightarrow p=bck,q=cak $ and $ r=-abk. $ Hence required plane is $ bcx+cay-abz=0 $ .
BETA
