Three Dimensional Geometry Question 33
Question: The point at which the line joining the points (2, -3, 1) and (3, -4, -5) intersects the plane $ 2x+y+z=7 $ is
[DSSE 1987; MP PET 1991]
Options:
A) (1, 2, 7)
B) (1, -2, 7)
C) (-1, 2, 7)
D) (1, -2, -7)
Show Answer
Answer:
Correct Answer: B
Solution:
Ratio $ -[ \frac{2(2)+(-,3)(1)+(1)(1)-7}{2(3)+(-,4)(1)+(-,5)(1)-7} ]=-[ \frac{-,5}{-,10} ]=-( \frac{1}{2} ) $
$ \therefore ,x=\frac{2,(2)-,3,(1)}{1}=1,y=\frac{-,3,(2),-,(,-,4)}{1}=-2 $ and $ z=\frac{1,(2),-,(-,5)}{1}=7 $ . Therefore, $ P,(1,,-,2,7) $ . Trick : As (1, ? 2, 7) and (? 1, 2, 7) satisfy the equation $ 2x+y+z=7, $ but the point (1, ? 2, 7) is collinear with (2, ? 3, 1) and (3, ? 4, ? 5). Note : If a point dividing the join of two points in some particular ratio, then this point must be collinear with the given points.
BETA
