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which fact is not sufficient to show that planes R and S are perpendicular?

FJ is contained in plane R, BC and DE are contained in plane S, and FJ, BC, and DE intersect at A.

its hard for me

6/17/2013 | Abel from Brooklyn, NY | 4 Answers | 0 Votes

Comments

The problem says:
1. FJ is contained in plane R.
2. BC and DE are contained in plane S.
3. Point A is contained in plane R and in plane S.
Rest is 
- maybe ...
- might be ...
- could be ...
In other words, are not sufficient facts.

6/17/2013 | Nataliya D.
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4 Answers

Just to add to the straight points made by my colleagues in Hallandale, FL and Franklin, TN ...

... I wish there were an easier way to make the points below ... Also, one has to strain their 3D imagination to see what I'm talking about ... please better employ right angle rulers and cardboard sheets ...

Making a long story short, plane R would be perpendicular onto plane S ... and vice-versa ...

... IF  {  { a 90° angle would exist between either FJ & BC or FJ & DE }  and  { of the 2 segments perpendicular to each other ... { one of the segments would be perpendicular onto the {R∩S} straight line } ... and ... { the other segment NOT being parallel/ included into the {R∩S} straight line }  }  } ...

... which tells us that ...

1. to assess 2 planes (R & S) as perpendicular onto each other, the value of some angles (in this case between any of the extensions of FJ, BC, DE intersecting in A) must be given ;

2. but such angles cannot be defined unless we know more about the orientation of FJ within plane R and the orientations of BC & DE within plane S ; specifically it is the orientation of these 3 segments defined with respect to the 2 plane {R∩S} straight line intersection that would make the difference.

I'm not sure if this helped.

7/3/2013 | Vladimir B.

Comments

Agree

Agree ......

In order for them to be perpendicular the vector normal to a plane must be parallel to the other plane

So either FJ have to be parallel to the normal vector at plane S

or ( BC or DE ) be parallel to the normal vector at plane R

7/3/2013 | Rafael V.

So far, we have only one fact: FJ, BC, and DE intersect at A. Other two, "FJ is contained in plane R, BC and DE are contained in plane S", are "subfacts" . 

The fact, that FJ, BC, and DE intersect at point A is not sufficient to show that planes R and S are perpendicular.
If we will know that FJ is perpendicular to BC or DE we will say that plane R   |   to plane S.

P.S. Abel, make sure you copy the problem completely.

6/17/2013 | Nataliya D.

Comments

If we will know that FJ is perpendicular to BC or DE we will say that plane R | to plane S.

I'm not so sure we can automatically conclude that R and S are perpendicular if FJ were perpendicular.

Based off the info provided in the problem we have to assume FJ is on R and BC and DE are on S. This however does not mean BC, DE, or FJ cannot be the line of intersection of the 2 planes. Say we let DE be the line of intersection (DE will still be on the plane S) then the line FJ could be perpendicular to DE and R not be perpendicular to S.

The only way to ensure the 2 planes are perpendicular is to show that a line on either plane is perpendicular to the other plane.

6/17/2013 | Xavier J.
Putting some of the foregoing in context: If FJ is perpendicular to both of two distinct (i.e. non-overlapping, but intersecting at A) segments in the other plane (which determine the second plane), then the planes are perpendicular. If not, then the conditions are insufficient to say (could be perpendicular, or not).
11/27/2013 | Stanton D.

As according to these two facts, R could rotate 360 degrees in both directions (Up/Down, Left/Right).  Does this help?

6/17/2013 | Rollin W.