No idea how to find the equation of a plane

[terryfields]

would this be ok for this question because i thought we had to find normals and things and every site i go on about plane equations is different and i need to know this for an exam tomorrow. find the equation of a plane that passes through the points (2,-2,1)(4,-1,6)(3,-3,-2)

(1)2a-2b+c=d
(2)4a-b+6c=d
(3)3a-3b-2c=d

then re arrange somehow to get values in terms of d for a b and c and then what? infact i have no clue what I am talking about i have no idea how to find the equation of a plane

 

[terryfields]

scrap that let's use the cross product methord using PQ=(0,1,-1) and PR=(-2,1,0) giving a normal vector of (1,2,2) after this it tells me

"(Q - P) x (R - P) = (1, 2, 2) = normal vector A and the equation is A . X = d for some d

Using the method in the example above, we can find d = A . P = 5. Thus the equation is A . X = 5, which is the same as one of the equations in the earlier example."

what the heck does this mean? i get everything up to finding the normal vector then it just all sounds forign

 

[terryfields]

hang on so does this mean that A.P=D and then we just stick the normal vector in as (A,B,C) using the formula Ax+By+Cz=D giving us an equation of x + 2y + 2Z=5 also if this is the case can someone check my one below that I am about to do

 

[rootX]

would this be ok for this question because i thought we had to find normals and things and every site i go on about plane equations is different and i need to know this for an exam tomorrow. find the equation of a plane that passes through the points (2,-2,1)(4,-1,6)(3,-3,-2)

(1)2a-2b+c=d
(2)4a-b+6c=d
(3)3a-3b-2c=d

then re arrange somehow to get values in terms of d for a b and c and then what? infact i have no clue what I am talking about i have no idea how to find the equation of a plane

1) find two direction vectors
e.g. (4,-1,6) - (2,-2,1)

2) find normal by taking cross product those two vectors

3) same problem as you asked before

 

[terryfields]

find eqn of plane through points (2,-2,1)(4,-1,6)(3,-3,-2)

normal is cross product of (b-a)(2,1,5)X(b-c)(1,2,8)=-2,11,3 so then we have N.A=-15 and we get -2x+11y+3z=-15??

 

[terryfields]

i know its the same as before but i still don't get it from before lol

 

[terryfields]

and ye your right they did just assume i knew this even though I've never studdied discrete maths

 

[rootX]

>> p1 = [2,-2,1];
>> p2 = [4,-1,6];
>> p3 = [3,-3,-2]; %pos vectors
>> d1 = p2-p1;
>> d2=p3-p2; %direction vectors
>> n = cross(d1,d2) %finding normal
n =

2 11 -3 %output
>> dot(n,p1)

ans =

-21

>> dot(n,p2)

ans =

-21

>> dot(n,p3)

ans =

-21=============
so everything is right but your d (above is MATLAB code ... - makes life easier - didn't have casio near which also do dots and cross)

 

[terryfields]

isnt it 2,-11,-3 and then when we get that do we do N.P1=D=23 so then the equation is 2x-11y-3z=23?

 

[rootX]

isnt it 2,-11,-3 and then when we get that do we do N.P1=D=23 so then the equation is 2x-11y-3z=23?

You must have done something wrong.

When we find n then

n dot any pos vector = should give equal value ..

>> n=[2,-11,-3];
>> dot(n,p1)

ans =

23

>> dot(n,p2)

ans =

1

>> dot(n,p3)

ans =

45

 

[terryfields]

ok ye just checked over it math error, apart from that tho is the rest of my working right subbing into get the Ax+By+Cz=D?