Proving Vectors B and C Act in the Same Direction When ( vec{A} vec{BC} ) and ( vec{A} vec{B} times vec{C} )

It can seem rather intuitive that vectors (vec{B}) and (vec{C}) must act in the same direction if (vec{A} vec{BC}) and (vec{A} vec{B} times vec{C}) holds true. However, we'll walk through a rigorous proof to ensure our understanding is sound.

Introduction to the Problem

Given the conditions: (vec{A} vec{BC}) and (vec{A} vec{B} times vec{C}). We need to prove that (vec{B}) and (vec{C}) must act in the same direction.

Step-by-Step Proof

Let's start by assuming that (vec{B}) makes an angle (alpha) with (vec{A}) and (vec{C}) makes an angle (beta) with (vec{A}).

Step 1: Component Analysis

The component of (vec{B}) in the (vec{A}) direction is (B cos alpha), and the component of (vec{C}) in the (vec{A}) direction is (C cos beta). Since (vec{A} vec{BC}), the components must align such that:

[ B cos alpha C cos beta ]

Given that (B cos alpha C cos beta A), we have:

[ cos alpha cos beta ]

The angles (alpha) and (beta) must thus be equal or differ by a multiple of (2pi), but since we're dealing with vectors in a standard geometric context, we can deduce that:

[ alpha beta 0 ]

This means that both (vec{B}) and (vec{C}) are acting in the same direction as (vec{A}).

Step 2: Using Dot Product

We can also use the dot product to solidify our proof. Let's denote the magnitudes of (vec{B}) and (vec{C}) as (B) and (C), and the angle between them as (theta).

Equation 1: Dot Product of (vec{A}) with itself

[ vec{A} cdot vec{A} A^2 BC cos 0 BC cdot 1 BC ]

Equation 2: Dot Product of (vec{B}vec{C}) with itself

[ vec{A} cdot vec{A} A^2 (vec{B} cdot vec{C})^2 B^2 (2 vec{B} cdot vec{C}) C^2 cos^2 theta ]

Setting the two expressions for (A^2) equal gives:

[ BC B^2 (2 vec{B} cdot vec{C}) C^2 cos^2 theta ]

Since (vec{B} cdot vec{C} BC cos theta), substitute to get:

[ BC B^2 (2 BC cos theta) C^2 cos^2 theta ]

This simplifies to:

[ BC 2 B^2 C^2 cos^3 theta ]

Given that (BC A^2), we can simplify to:

[ A^2 2 B^2 C^2 cos^3 theta ]

Since (A^2 B^2 C^2 cos^2 theta) from (vec{A} vec{B} times vec{C}), we have:

[ cos^2 theta 1 ]

Hence:

[ cos theta 1 ]

This means that the angle between (vec{B}) and (vec{C}) is zero, confirming that they act in the same direction.

Geometric Interpretation

This can be visualized geometrically by using the parallelogram rule for vector addition. In a parallelogram formed by (vec{B}) and (vec{C}), the only way for the diagonal ((vec{A})) to be the sum of the lengths of the adjacent sides is for the angle between them to be zero.

Conclusion

Both the algebraic and geometric approaches confirm that (vec{B}) and (vec{C}) must act in the same direction when (vec{A} vec{BC}) and (vec{A} vec{B} times vec{C}).