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Question:For example, my homework last night said "determine if the points are co planer". It showed a picture of a quadrilateral (not a cube) with random points on and off of it, some on lines and some not. I know that two points are always coplaner and three non collinear points are always co planer, but four points are only sometimes coplaner. So how do you tell when the four points (on a flat diagram of a quadrilateral) are coplaner? :/
Answers:Pick three of the points and form a triangle with them, they are coplanar so this defines a plane. You want to know if the fourth point lies in this plane or not. This is a very standard procedure using vectors; I recall there is a formula for determining the minimum distance of a point from a plane. So if this distance is zero then it must lie in the plane. I can't quite remember my basic school maths, I'm hoping you will have seen it, I'll will post it if I find it. Edit: Aha! I remember now. It's VERY easy to do. Ignore what I said before, don't bother with planes and 'normal vectors'; that just gets messy. You just pick a point call it A. And you calculate the volume of the parallelapiped formed by the vectors AB, AC and AD. If this volume is zero, then they must be coplanar. To work out the volume, you just take the scalartriple product of the three vectors  piece of cake. I hope you know how to do the scalartripleproduct but look it up if you haven't, it's just a simple function (just adding and multiplying some of the coordinates) which takes three vectors and gives you a number, which we hope will come out to be 0.
Answers:Pick three of the points and form a triangle with them, they are coplanar so this defines a plane. You want to know if the fourth point lies in this plane or not. This is a very standard procedure using vectors; I recall there is a formula for determining the minimum distance of a point from a plane. So if this distance is zero then it must lie in the plane. I can't quite remember my basic school maths, I'm hoping you will have seen it, I'll will post it if I find it. Edit: Aha! I remember now. It's VERY easy to do. Ignore what I said before, don't bother with planes and 'normal vectors'; that just gets messy. You just pick a point call it A. And you calculate the volume of the parallelapiped formed by the vectors AB, AC and AD. If this volume is zero, then they must be coplanar. To work out the volume, you just take the scalartriple product of the three vectors  piece of cake. I hope you know how to do the scalartripleproduct but look it up if you haven't, it's just a simple function (just adding and multiplying some of the coordinates) which takes three vectors and gives you a number, which we hope will come out to be 0.
Question:i already know that if they r in the same plane they r coplanar. but what if 2 points r not in the plane or not in the same plane? How can i tell if they r coplanar? My teacher said if u can draw plane around them... but to me, it sometimes doesn't work.
Answers:There isn't much to go on here. Three points determine a plane so if you have such a plane and are given a 4th point and want to know if it lies in the plane, one way to do that would be to calculate the volume of a solid formed by these points. If the volume is 0 then the point is in the plane. If A, B, C and D are the points. D is the point to determine if it lies in the plane. AB = vector from A to B AC = vector from A to C AD = vector from A to D The volume is: AD . (AB x AC) where "." is the dot product and "x" is the cross product. The cross product gives you a vector perpendicular to the plane and whose magnitude is the area in the plane of a parallelogram formed using AB and AC. When you take the dot product using AD this is then the height of a solid times the area of it's base (dot product is the magnitude of each vector times the cosine of the angle between them). If AB . (AC x AD) = 0 then D lies in the plane since there is no sold. Is this the sort of answer you are looking for?
Answers:There isn't much to go on here. Three points determine a plane so if you have such a plane and are given a 4th point and want to know if it lies in the plane, one way to do that would be to calculate the volume of a solid formed by these points. If the volume is 0 then the point is in the plane. If A, B, C and D are the points. D is the point to determine if it lies in the plane. AB = vector from A to B AC = vector from A to C AD = vector from A to D The volume is: AD . (AB x AC) where "." is the dot product and "x" is the cross product. The cross product gives you a vector perpendicular to the plane and whose magnitude is the area in the plane of a parallelogram formed using AB and AC. When you take the dot product using AD this is then the height of a solid times the area of it's base (dot product is the magnitude of each vector times the cosine of the angle between them). If AB . (AC x AD) = 0 then D lies in the plane since there is no sold. Is this the sort of answer you are looking for?
Question:Okay, how do i draw 4 points A,B,C,D as collinear and coplanar?
Answers:just draw a line with those points on it and then make it look like a square or something lol.
Answers:just draw a line with those points on it and then make it look like a square or something lol.
Question:Consider the four points (1,0,0) (2,0,x) (1,x,0) (1+x,x,x). For what values of x are the points coplanar?
Answers:Find the 3 vectors that link each of the points to the point (1,0,0) e.g. for (2,0,x) this would be (1,0,x) if all of these points are in coplanar, the vectors will lie in the same plane. Taking the dot product of any of these vectors with the normal vector n = (a,b,c) of the plane will result in a value of zero, as the vectors are all perpendicular to the normal vector. So you can set up a matrix equation with each of the vectors as a row of the matrix. (vec 1) (a) (0) (vec 2) (b) = (0) (vec 3) (c) (0) Finding the determinant of the matrix and setting this to zero will give you all values of x for the points to keep them coplanar  I think there are only 2 values...
Answers:Find the 3 vectors that link each of the points to the point (1,0,0) e.g. for (2,0,x) this would be (1,0,x) if all of these points are in coplanar, the vectors will lie in the same plane. Taking the dot product of any of these vectors with the normal vector n = (a,b,c) of the plane will result in a value of zero, as the vectors are all perpendicular to the normal vector. So you can set up a matrix equation with each of the vectors as a row of the matrix. (vec 1) (a) (0) (vec 2) (b) = (0) (vec 3) (c) (0) Finding the determinant of the matrix and setting this to zero will give you all values of x for the points to keep them coplanar  I think there are only 2 values...