Q.$\vec{A}$, $\vec{B}$ and $\vec{C}$ are three non-collinear, non co-planar vectors. What can you say about direction of $\vec{A} \times (\vec{B} \times \vec{C})$?

The key idea is the vector triple product identity:

Since $\vec{A}$, $\vec{B}$, $\vec{C}$ are non-collinear and non-coplanar, the scalars $(\vec{A} \cdot \vec{C})$ and $(\vec{A} \cdot \vec{B})$ are not both zero. The result is a linear combination of $\vec{B}$ and $\vec{C}$. …

The key to this question is understanding what a cross product physically means. When you take $\vec{B} \times \vec{C}$, you get a vector perpendicular to both $\vec{B}$ and $\vec{C}$. Then crossing that result with $\vec{A}$ gives a vector perpendicular to both $\vec{A}$ and $\vec{B} \times \vec{C}$. But since $\vec{B} \times \vec{C}$ is perpendicular to the plane of $\vec{B}$ and $\vec{C}$, any vector perpendicular to $\vec{B} \times \vec{C}$ must lie in that same plane. So the final vector is forced into the plane of $\vec{B}$ and $\vec{C}$.

Let’s walk through this carefully.

1. First, the inner cross product: $\vec{B} \times \vec{C}$ is a vector perpendicular to both $\vec{B}$ and $\vec{C}$. Since $\vec{B}$ and $\vec{C}$ are non-collinear, they define a unique plane. The cross product points along the normal to that plane (direction given by the right-hand rule).

2. Now the outer cross product: $\vec{A} \times (\vec{B} \times \vec{C})$ is perpendicular to both $\vec{A}$ and $\vec{B} \times \vec{C}$. Being perpendicular to $\vec{B} \times \vec{C}$ means it lies in the plane that $\vec{B} \times \vec{C}$ is normal to — which is exactly the plane of $\vec{B}$ and $\vec{C}$.

3. So the result is simultaneously:

   • In the plane of $\vec{B}$ and $\vec{C}$ (because it’s perpendicular to $\vec{B} \times \vec{C}$).
   • Perpendicular to $\vec{A}$ (because it’s the cross product with $\vec{A}$).

4. What about direction within that plane? The vector triple product identity gives the explicit expression:

   $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C}$ …

Concept: Choose Coordinates Aligned with $\vec B,\vec C$ and Watch the $z$-Component Vanish

Method: Adapted Coordinate Frame (Put $\vec B,\vec C$ in the $xy$-plane), Not the BAC-CAB Identity

Rather than quoting the vector triple-product identity $\vec A\times(\vec B\times\vec C) = (\vec A\cdot\vec C)\vec B - (\vec A\cdot\vec B)\vec C$ as a known formula, this method builds a coordinate frame specifically adapted to $\vec B$ and $\vec C$ -- placing their entire plane along $xy$ -- and shows the result must lie in that plane by watching one component vanish identically in the computation itself.

Step 1 -- Adapt the axes to $\vec B$ and $\vec C$

Since $\vec B,\vec C$ are non-collinear, they span a plane. Choose axes so that this plane is the $xy$-plane: let $\hat x$ point along $\vec B$, and let $\vec C$ have components only in $x$ and $y$ (no $z$). Explicitly:

Since $\vec A,\vec B,\vec C$ are non-coplanar, $\vec A$ must have a nonzero $z$-component in this frame (otherwise it would lie in the $\vec B,\vec C$ plane, contradicting non-coplanarity): $\vec A = A_x\hat i + A_y\hat j + A_z\hat k$, with $A_z\neq0$.

Step 2 -- Compute the inner cross product $\vec B\times\vec C$ in this frame

Because $\vec B,\vec C$ were placed entirely in the $xy$-plane, their cross product points purely along $\hat k$ -- confirming the general fact that $\vec B\times\vec C$ is normal to the plane of $\vec B,\vec C$, but now as an explicit component computation rather than an assertion.

Step 3 -- Compute the outer cross product $\vec A\times(\vec B\times\vec C)$ component by component

Using $\hat i\times\hat k=-\hat j$, $\hat j\times\hat k=\hat i$, $\hat k\times\hat k=\vec 0$:

Step 4 -- Read the answer directly: the $\hat k$-component is exactly zero

The result $BC_y(A_y\hat i - A_x\hat j)$ has no $\hat k$-component at all -- notice that $A_z$, which was the only nonzero indicator that $\vec A$ points out of the $\vec B,\vec C$ plane, has completely dropped out of the final expression. This is not a coincidence of this particular frame -- the $\hat k\times\hat k=\vec 0$ step in Step 3 is exactly what kills any contribution from $\vec A$'s out-of-plane component. …