Three Dimensional Geometry Question 161
Question: The point where the line $ \frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4} $ meets the plane $ 2x+4y-z=1 $ , is
[DSSE 1981]
Options:
A) (3, ?1, 1)
B) (3, 1, 1)
C) (1, 1, 3)
D) (1, 3, 1)
Show Answer
Answer:
Correct Answer: A
Solution:
Let point be (a, b, c), then $ 2a+4b-c=1 $ …..(i) and $ a=2k+1,b=-3k+2 $ and $ c=4k-3 $ , (where k is constant) Substituting these values in (i), we get $ 2,(2k+1)+4,(-3k+2)-(4k-3)=1\Rightarrow k=1 $ Hence required point is (3, ?1, 1). Trick : The point must satisfy the lines and plane. Obviously (3, ? 1, 1) satisfies.
BETA
