BAC - CAB Special Case

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Let vectors $\vec A, \vec B$ represent the sides of a quadrilateral parallelogram.

$\vec B \times (\vec A \times \vec B) $

$= \vec A(\vec B \cdot \vec B) - \vec B(\vec B \cdot \vec A)$

$= B^2 \vec A - \vec B(\vec B \cdot \vec A)$

Is the last line typed above the simplest way to express the triple cross product when two of the three vectors are the same?

Or can I simplify it to:

$= B^2 \vec A - (\vec B \vec B \cdot \vec B \vec A)$

Please point me in the right direction. I could not find this "vector property" aka:

vector times ( vector dot vector)

anywhere online.