maliaseth0

2022-01-24

How could I determine whether vectors $P<-2,7,4>,Q<-4,8,1>,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}R<0,6,7>$ are all in the same plane?

Mazzuranavf

Expert

The first way is to use linear algebra. If a matrix is dependent, then the vectors are coplanar. So, you need to try reducing the matrix:
$\left[\begin{array}{ccc}-2& 7& 4\\ -4& 8& 1\\ 0& 6& 7\end{array}\right]$
$\left[\begin{array}{ccc}-2& 7& 4\\ 0& -6& -7\\ 0& 6& 7\end{array}\right]{R}_{2}-2{R}_{1}$
$\left[\begin{array}{ccc}-2& 7& 4\\ 0& -6& -7\\ 0& 0& 0\end{array}\right]{R}_{3}+{R}_{2}$
The third row is zero, so the matrix is dependent and therefore the vectors are coplanar.
The second way is to use the scalar triple product. One interpretation of this product is that it is the volume of a parallelepiped formed by the vectors. If the volume is zero, then the vectors must be coplanar. Here is a simplification of the scalar triple product:
$A\cdot \left(B×C\right)$
$=A\cdot \left[\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ {b}_{1}& {b}_{2}& {b}_{3}\\ {c}_{1}& {c}_{2}& {c}_{3}\end{array}\right]$
$=\left[\begin{array}{ccc}{a}_{1}& {a}_{2}& {a}_{3}\\ {b}_{1}& {b}_{2}& {b}_{3}\\ {c}_{1}& {c}_{2}& {c}_{3}\end{array}\right]=|M|$
$=det|M|$
So, we can just find the determinant of the vectors. If the determinant (volume of the parallelepiped) is zero, then the vectors are coplanar:
$\left[\begin{array}{ccc}-2& 7& 4\\ -4& 8& 1\\ 0& 6& 7\end{array}\right]$
$=-2\left[8\left(7\right)-1\left(6\right)\right]-7\left[-4\left(7\right)-1\left(0\right)\right]+4\left[-4\left(6\right)-8\left(0\right)\right]$
$=-2\left[50\right]-7\left[-28\right]+4\left[-24\right]$
$=-100+196-96$
$=0$
And we do have a 0 determinant, so they are coplanar, again.

Bottisiooq

Expert

If this is all hard to wrap your head around, you can do it the long way. The cross product will find the normal between 2 vectors. So compute the normal between 2 of the vectors, say ${N}_{1}=P×R$ then do it for another 2 vectors, say ${N}_{2}=Q×R$. If all 3 are coplanar, then ${N}_{1}×{N}_{2}$ or $\left(P×R\right)×\left(Q×R\right)$ will be zero. I'm sure there are some identities that will convert this close to the scalar triple product!