Three Dimensional Geometry Question 198
Question: The line joining the points (3, 5, ?7) and (?2, 1, 8) meets the yz-plane at point
[RPET 2003]
Options:
A) $ ( 0,,\frac{13}{5},,2 ) $
B) $ ( 2,,0,,\frac{13}{5} ) $
C) $ ( 0,,2,,\frac{13}{5} ) $
D) (2, 2, 0)
Show Answer
Answer:
Correct Answer: A
Solution:
Line joining the points (3,5,?7) and (?2,1,8) is, $ \frac{x-3}{(-2)-(3)}=\frac{y-5}{(1)-(5)}=\frac{z-(-7)}{8-(-7)} $ $ \frac{x-3}{-5}=\frac{y-5}{-4}=\frac{z+7}{15}=K $ , (Let) ?..(i) \ $ x=-5K+3 $ , $ y=-4K+5 $ , $ z=15K-7 $ $ \because $ Line (i) meets the yz-plane
$ \therefore $ $ -5K+3=0\Rightarrow K=3/5 $ Put the value of K in $ x,,y,,z $ So the required point is (0, 13/5, 2).
BETA
