IIT JEE , AIEEE Forum - Mathematics , Vectors

masterimp

find the no. of planes that are equidistant from four non-coplanar points.

rashmi_jain

take any three points,they all will lie on a plane(say plane P)

now,there will be one plane(say plane A) which is equidistant from P and the 4th point.

also,since first three points lie on P,distance of those points from A will be the same as distance of fourth point from A.

since there are 4 ways of choosing these three points,we will get 4 different planes which are equidistant from four non coplanar points.

masterimp

wrong!! the answer is 7 (ref. fiitjee aits) and i cannot make the head or tail of it (how to get 7)

rebel_abhi

yes even i m stuck wid dis ques....i also chose 4 with d same logic n got it wrong!!!!

hit_ur_heart

Friends let me write the solution of this problem and also the mistake in the solution provided by rashmi

What rashmi has done is that she has taken 3 points and then she draws a plane passing through all those points and then taken 4th point and she says we will find a plane which is equidistant from the plane we have drawn and from the last point ...........................this is correct , but is it exhaustive ......................no definitely not
What about those cases in which plane will separate four points , two above it and two below it...................................definitely that case isnt covered , and there is
4C2/2 ways to do this(divided by two because planes will be same in half of the case) , i.e , 3 cases.
Total number of possible planes are 7

This is correct because it is exhaustive , means we cannot choose 4 points on same , plane as that will violate their non-coplanarity and 3,1 and 1,3 case are same , and (2,2) case is now covered.

rahul_c

4 non coplanarpoints can be arranged in space 4!=24 ways

WE HAVE TO ELEMINATE THE CASES WHEN THEY BECOME COPLANAR

say if 2 points becomes co-planar which can be chosen in 4C2 ways ,remaining 2 points can be arranged in 2! ways
so total mo.of ways=4C2*2!=12

say if 3 points becomes co-planar which can be chosen in 4C3 ways ,remaining 1 point can be arranged in 1! ways
so total mo.of ways=4C3*1!=4

now if all 4 points become coplanar which can be chosen in 4C4 ways =1

therefore total number of ways=4!-(4C2*21 + 4C3*1 + 4C4)=7 WAYS

with regards

hit_ur_heart

rahul , ur solution is wrong

Do u have any idea what u r doing in the answer

In question it is given that , points are non-coplanar and u r trying to find cases when points are non-coplanar, well dude there are infinite number of ways in which these points can be non-coplanar(think over it why), and next on what basis u r finding ways , how u r arranging these points ........................................................................................
And last fact , u r trying to say that given 4 points whenever u will make them non-coplanar they will give u a plane in between them which is equidistant from all four points ..............................i.e , completely irrelevant

Last but not the least question says , find the planes equidistant from 4 non-coplanar points that means points are fixed in space , and only information we are having about them is that they are coplanar , nothing more than that

rahul_c

i am sure of the answer.
when points are arranged in space,how do you know they are noncoplanar?
it is by eleminating the conditons f coplanarityi

rebel_abhi

saluting hit ur heart's answer nd brains...

Ask & Discuss Questions with Community & Experts