@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