Three Dimensional Geometry Question 313
Question: Image point of $ (1,,3,4) $ in the plane $ 2x-y+z+3=0 $ is
Options:
A) (? 3, 5, 2)
B) (3, 5, ? 2)
C) (3, ? 5, 3)
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let Q be image of the point $ P(1,3,,4) $ in the given plane, then PQ is normal to the plane. The d.r.’s of PQ are 2, ?1,1 Since PQ passes through (1, 3, 4) and has d.r’s 2, 1, ?1; therefore, equation of plane is $ \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=r $ , (say)
$ \therefore $ $ x=2r+1,,y=-r+3,,z=r+4 $ So, co-ordinates of Q be $ (2r+1,,-r+3,,r+4) $ Let R be the mid point of PQ, then co-ordiantes of R are $ ( ,\frac{2r+1+1}{2},,\frac{-r+3+3}{2},,\frac{r+4+4}{2} ) $ i.e., $ ( r+1,,\frac{-r+6}{2},,\frac{r+8}{2} ) $ Since R lies on the plane
$ \therefore $ $ 2(r+1)-( \frac{-r+6}{2} )+( \frac{r+8}{2} )+3=0 $
Þ $ r=-2 $ So, co-ordinates of Q are (?3, 5, 2). Trick : From option , mid point of (?3, 5, 2) and (1,3,4) satisfies the equation of plane $ 2x-y+z+3=0 $ .
BETA
