Question

Determine if u= [2, 7, -1], v= [5, -3, 2] and w= [9, 11, 2] are coplanar.

Answer 1

are those points or vectors?

Answer 2

if vectors; then by adding or subtracting 2 together you chould be able to produce the 3rd

Answer 3

vectors

Answer 4

2, 7, -1
5, -3, 2
-------
7, -4, 1 ..... try again?



2, 7, -1
-5, 3, -2
---------
-3,10,-3 .... well those dont pan out yet

Answer 5

i spose we could start them at a point say(0,0,0) and see if they end up back at it?

Answer 6

co planar does mean all on the same plane right?

Answer 7

yes

Answer 8

u= [2, 7, -1], v= [5, -3, 2] and w= [9, 11, 2]

(0 ,0, 0)
<2,7,-1>
----------
(2,7,-1)
<5,-3,2>
----------
(7,4,1)
<9,11,2>
----------
(16,15,3) we dont get back to the same point ..... but thats not definitive is it, since the vectors could just be short, as is

Answer 9

(0 ,0, 0)
<2,7,-1>
----------
(2,7,-1)
<5,-3,2>
----------
(7,4,1)
<9,11,2> ; if we can scale this vector to get back to (0,0,0)
---------- that should be good; but I dont see it happening.

My gut says they aint co planar

Answer 10

what have you tried?

Answer 11

i did the long way

Answer 12

so with s and t

Answer 13

and it doesnt equal

Answer 14

the long way is fine :)

Answer 15

I considered last night crossing each vector: u(x)v u(x)w and v(x)w if they all produce the same "normal" vector to thier plane; then they would be coplanar.

u [2, 7, -1]
v [5, -3, 2]
------------
(7.2)-(-3.-1),(-1.5)-(2.2),(2.-3)-(5.7)
14 - 3 , -5 - 4 , -6 - 35
11 , -9 , -41


u [2, 7, -1]
w [9, 11, 2]
------------
(7.2)-(11.-1),(9.-1)-(2.2),(2.2)-(9.-1)
14 +11 , -9-4 , 4+9
25 , -13 , -13

v [5,-3,2]
w[9,11,2]
-------------
(-3.2)-(11.2),(9.2)-(5.2),(5.11)-(9.-3)
-6 - 22 , 18-10 , 55+27
-28 , 8 , 82


the "normals" produced by crossing are:

<11,-9,-41>
<25,-13,-13>
<-28, 8 , 82>

And I see no way that these could even be remotely the same. Since all 3 crossings produce different "normal" vectors; they cannot be coplanar.