Solve for Area with Given Vertices Using Vector Calculus

[matt222]

Homework Statement

find the area if the vertices are (3,9,8),(0,5,1),(-1,-3,-3),(2,1,4)

Homework Equations

The Attempt at a Solution

I draw the points and I couldn't know the shape it is complex I really couldn't know it

 

[tiny-tim]

hi matt222!

find the area if the vertices are (3,9,8),(0,5,1),(-1,-3,-3),(2,1,4)

do you mean the volume?

 

[zgozvrm]

To find the volume, see http://www.jtaylor1142001.net/calcjat/Solutions/VCrossProduct/VCP4PtsCoplanar.htm.

If the result is not 0, you have the volume.

If the result is 0, the points are coplanar and there may be several ways to connect the points to create a closed shape of which we can determine the area.

 

[matt222]

its equal to zero and its indeed coplanar, but how to get the area?

 

[matt222]

I draw them but they are not perfectly clear I tried to connect the point with different ways but I have many answers which in point of view not true, is there any rules to get the area

 

[zgozvrm]

I draw them but they are not perfectly clear I tried to connect the point with different ways but I have many answers which in point of view not true, is there any rules to get the area

... and therein lies the problem!

With the information given, there is no way to determine how to "connect the dots," so I assume the original question was to determine the volume, rather than the area (which you have already done).

 

[tiny-tim]

there's only two possibilities (if they're coplanar) …

they form a convex quadrilateral, or one point is "inside" the other three

you could find the area of the four triangles …

if three add to make the fourth, then it's a convex quadrilateral, and the area of the fourth is the total area

if two add to make the same sum as the other two, then that sum is the area

(alternatively, i expect there's a way of assigning a sign to the area of each triangle which will actually tell you the layout)

 

[zgozvrm]

there's only two possibilities (if they're coplanar) …

they form a convex quadrilateral, or one point is "inside" the other three

you could find the area of the four triangles …

if three add to make the fourth, then it's a convex quadrilateral, and the area of the fourth is the total area

if two add to make the same sum as the other two, then that sum is the area

(alternatively, i expect there's a way of assigning a sign to the area of each triangle which will actually tell you the layout)

If one point is "inside" the other three, then the sum of the areas of the small triangles (the 3 triangles that contain that "inside" point) will be equal to the area of the "outside" triangle.

We could assume that the area in question is the area of the "outside" triangle (basically ignoring the "inside" point altogether). However, if you look at my first attachment (Inside.JPG), you'll see that there are 3 other possible scenarios for connecting those same 4 points. And it is entirely possible that the area of each of them is different than that of the others (in fact, in my picture, they are different).


On the other hand, if no point is "inside" the other 3, you have a quadrilateral which has an area equal to the sum of the 4 triangles produced by the 4 points. But, without more information, there is nothing telling us that the points can't be connected as in my 2nd attachment (Outside.JPG). In these 2 cases, you have only 2 triangles.