How can I prove that a set of vectors belong to the same plane given three of them?

Jun 2017
352
6
Lima, Peru
The problem is as follows:

Vectors $\vec{A}=2\hat{i}-\hat{j}+\vec{k}$, $\vec{B}=\hat{i}+2\hat{j}-3\hat{k}$ and $\vec{C}=3\hat{i}+a\hat{j}+5\hat{k}$. Find the value of $a$ such the vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ belong to the same plane.

The alternatives are as follows:

$\begin{array}{ll}
1.&2\\
2.&-1\\
3.&4\\
4.&-4\\
5.&-2\\
\end{array}$

I'm not exactly sure which sort of property or identity should be used here. But I'm guessing that it is has to do that
I have to define the plane and use the cross product. But I don't know exactly how to tie this with the first. Can somebody help me here?.
 

romsek

Math Team
Sep 2015
2,891
1,615
USA
if the three vectors are coplanar the triple product will be zero.

$A\cdot (B\times C)=0$

$A=(2,-1,1),~B=(1,2,-3),~C=(3,a,5)$

$B\times C = (10 + 3 a, -14, -6 + a)$

$A \cdot (B \times C) = 28+7a = 0 \Rightarrow a = -4$

Another way of looking at this is you can form a 3x3 matrix using the vectors.
If those vectors are coplanar then the determinant of this matrix will be zero indicating that the system isn't full rank.

The operations of finding this determinant are identical to those of taking the triple product.
 
Jun 2017
352
6
Lima, Peru
if the three vectors are coplanar the triple product will be zero.

$A\cdot (B\times C)=0$

$A=(2,-1,1),~B=(1,2,-3),~C=(3,a,5)$

$B\times C = (10 + 3 a, -14, -6 + a)$

$A \cdot (B \times C) = 28+7a = 0 \Rightarrow a = -4$

Another way of looking at this is you can form a 3x3 matrix using the vectors.
If those vectors are coplanar then the determinant of this matrix will be zero indicating that the system isn't full rank.

The operations of finding this determinant are identical to those of taking the triple product.
In your solution you used that:

$A\cdot (B\times C)=0$

But I'm confused at what were you trying to say with

... triple product will be zero..
Did you intend to say this product?

$A \times B\times C=0$

Or this product?

$A \cdot B\cdot C=0$

Why the first has to be a dot product and the other has to be a cross product?. Does it exist a justification for this?. Sorry for that, but it is not too obvious for me.
 

topsquark

Math Team
May 2013
2,452
1,017
The Astral plane
\(\displaystyle \vec{A} \times \vec{B} \times \vec{C}\) can be calculated but \(\displaystyle \vec{A} \cdot \vec{B} \cdot \vec{C}\) doesn't exist. \(\displaystyle \vec{A} \cdot \vec{B}\) gives you a scalar quantity and you can't "dot" a vector and scalar.

There was no misprint: \(\displaystyle \vec{A} \cdot ( \vec{B} \times \vec{C} )\) is not a typo. It's is called the "triple product" or sometimes the "box product." See here.

-Dan
 
Jun 2017
352
6
Lima, Peru
\(\displaystyle \vec{A} \times \vec{B} \times \vec{C}\) can be calculated but \(\displaystyle \vec{A} \cdot \vec{B} \cdot \vec{C}\) doesn't exist. \(\displaystyle \vec{A} \cdot \vec{B}\) gives you a scalar quantity and you can't "dot" a vector and scalar.

There was no misprint: \(\displaystyle \vec{A} \cdot ( \vec{B} \times \vec{C} )\) is not a typo. It's is called the "triple product" or sometimes the "box product." See here.

-Dan
@topsquark Thanks Dan. Your clarification was much needed. Out of curiosity, how would you calculate $\vec{A} \times \vec{B} \times \vec{C}$?. And what's the justification that the triple product must be zero?. Does it exist some proof or visual explanation for this?. I would like to know the reason why does this approach works in my problem.
 

topsquark

Math Team
May 2013
2,452
1,017
The Astral plane
@topsquark Thanks Dan. Your clarification was much needed. Out of curiosity, how would you calculate $\vec{A} \times \vec{B} \times \vec{C}$?. And what's the justification that the triple product must be zero?. Does it exist some proof or visual explanation for this?. I would like to know the reason why does this approach works in my problem.
I am going to take \(\displaystyle \vec{A} \times \vec{B} \times \vec{C}\) as \(\displaystyle \vec{A} \times ( \vec{B} \times \vec{C} )\)
(The cross product is not commutative so order matters.)

\(\displaystyle \vec{B} \times \vec{C} = \left | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{matrix} \right |\)
where |,| is the determiant of the matrix. Let \(\displaystyle \vec{D} =\left | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{matrix} \right | = D_x \hat{i} + D_y \hat{j} + D_z \hat{k}\)

Then \(\displaystyle \vec{A} \times ( \vec{B} \times \vec{C} ) = \vec{A} \times \vec{D} = \left | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ D_x & D_y & D_z \end{matrix} \right | \)
You could write out each component but it doesn't get simpler than this.

Now, there is another way to write out \(\displaystyle \vec{A} \times ( \vec{B} \times \vec{C} ) = (\vec{A} \cdot \vec{C} ) \vec{B} - ( \vec{A} \cdot \vec{B} ) \vec{C}\)
which comes in handy at times, but I'm betting that's not what you need here.

However triple product is pretty nice:
\(\displaystyle \vec{A} \cdot ( \vec{B} \times \vec{C} )= \vec{A} \cdot \left | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{matrix} \right | = \left | \begin{matrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{matrix} \right |\)
which is nice and simple to remember.

Now, let \(\displaystyle \vec{B}\) and \(\displaystyle \vec{C}\) form a plane. Then \(\displaystyle \vec{D} = \vec{B} \times \vec{C}\) is perpendicular to \(\displaystyle \vec{A}\) (because \(\displaystyle \vec{A}\) is in the same plane formed by \(\displaystyle \vec{B}\) and \(\displaystyle \vec{C}\) ) and the dot product between two perpendular vectors is 0. Thus \(\displaystyle \vec{A} \cdot ( \vec{B} \times \vec{C} ) = 0\) for this case.

-Dan