# Verify Vectors are Coplanar w/o Scalar Triple Product

• Dec 16th 2010, 06:27 PM
divinelogos
Verify Vectors are Coplanar w/o Scalar Triple Product
Problem:

Verify that the vectors <2,3,1>, <1,-1,0>, and <7,3,2> are coplanar.

Question:

Of course, the easiest way to do this is using the scalar triple product. At the time of my test, however, I was unaware of this. I attempted to find the angle between the vectors, and I got the result that the angle did not exist for any vectors I tried. So, if there is no angle between any of the vectors, does it follow that they are coplanar?
• Dec 16th 2010, 06:28 PM
dwsmith
Take the determinant. If the determinant is 0, the vectors are linearly dependent, i.e. coplanar.

$\displaystyle
\begin{vmatrix}
2 & 1 & 7 \\
3 & -1 & 3 \\
1 & 0 & 2
\end{vmatrix}=0 \ \mbox{or} \ rref=\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 3 \\
0 & 0 & 0
\end{bmatrix}$

Therefore, $2<2,3,1>+3<1,-1,0>=<7,3,2>$
• Dec 16th 2010, 06:54 PM
divinelogos
I know how to do it using the scalar triple product, my real question is this though:

If you use the standard formula for finding the angle between the vectors, and come out with "Does not exist", does it follow that the vectors are coplanar?
• Dec 16th 2010, 06:59 PM
snowtea
The angle always exists between two nonzero vectors.
v dot w = |v||w|cos(angle)
How can the angle be undefined? (Unless one of |v| or |w| is 0, but that is not the case for this problem)

Also, the angle won't help with being coplanar.
Think about 3 vectors in the xy-plane. They have to be coplanar, but there is no relation between their angles.
• Dec 16th 2010, 06:59 PM
dwsmith
Counter example

u=<1,1>, v=<1,1>

$\displaystyle
\theta=cos^{-1}\left(\frac{}{||u|| \ ||v||}\right)=cos^{-1}\left(\frac{2}{2}\right)=cos^{-1}(1)=0$

Therefore, that isn't true since these two coplanar vectors have an angle of 0
• Dec 17th 2010, 02:03 AM
HallsofIvy
Quote:

Originally Posted by divinelogos
I know how to do it using the scalar triple product, my real question is this though:

If you use the standard formula for finding the angle between the vectors, and come out with "Does not exist", does it follow that the vectors are coplanar?

Angle between what two vectors? The angle between two non-zero vectors always exists!

You could, as dwsmith did at the end of his post, using the result from the post, show that one of the vectors is a linear combination of the other two. That is, <2,3,1>, <1,-1,0>, and <7,3,2> are co-planar if and only if there exist numbers a and b such that <7, 3, 2>= a<2, 3, 1>+ b<1, -1, 0>. That is the same as saying <7, 3, 2>= <2a+ b, 3a- b, a> or the system of equations 2a+ b= 7, 3a- b= 3, a= 2. Since the last equation tells us immediately that a= 2, the first equation becomes 4+ b= 7 or b= 3. Putting a= 2 and b= 3 in the second equation we get 3(2)- 3= 6- 3= 3 so they also satisfy the second equation. That tells us that the three vectors are co-planar.
• Dec 17th 2010, 09:19 AM
SammyS
Quote:

Originally Posted by divinelogos
Problem:

Verify that the vectors <2,3,1>, <1,-1,0>, and <7,3,2> are coplanar.

Question:

Of course, the easiest way to do this is using the scalar triple product. At the time of my test, however, I was unaware of this. I attempted to find the angle between the vectors, and I got the result that the angle did not exist for any vectors I tried. So, if there is no angle between any of the vectors, does it follow that they are coplanar?

Call the three vectors $\displaystyle \vec{a},\ \vec{b},\ \vec{c}\ .$

Use the scalar product to find the angle between each pair of vectors.

The angle between $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ is: $\displaystyle \theta_{ab}=\cos^{-1}\left({{\vec{a} \cdot \vec{b}}\over{|\vec{a}|\ |\vec{b}|}}\right).$

Find $\displaystyle \theta_{bc}$ and $\displaystyle \theta_{ac}$ similarly.

If the vectors are co-planar, then one of the two following things will be true.
1. The sum of two of these angles is equal to the third.

2. The sum of all three angles will be 360° (or $\displaystyle 2\pi$ radians).
For the three given angles, #1 works.

You could do a similar process using the vector product. Then $\displaystyle \theta_{ab}=\sin^{-1}\left({{|\vec{a} \times \vec{b}|}\over{|\vec{a}|\ |\vec{b}|}}\right).$

In this case, criterion #2 would change to: The sum of the three angles being 180°, rather than 360°, would indicate that the vectors are co-planar.