# Is back of the cab constant?

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

which is also called "back of the cab". I know that cross product is vector, like as $$\vec C=\vec A\times \vec B$$ so C is vector. But when I used dot product they are forming a scalar. So my question is do we really use the expression (back of the cab) for magnitude of triple cross product?

$$C=\vec A\cdot \vec B$$

Editing again after seeing Jean's comment (first comment) :

When I used dot product that became scalar as I said earlier so I am going to assume

$$A\cdot B=\alpha \\ C\cdot A=\beta$$

alpha and beta is scalar now but A and B is scalar or vector, who cares? When I multiplied alpha and beta by B respectively with B then I got

$$B\cdot \alpha-\beta \cdot B$$ Now they are scalar. Who says they are vector?

• Good mnemonic indeed ! As your pseudo, I ignored it. Besides, the result is not a magnitude (which is a positive number) but a vector. Nov 14, 2021 at 10:36
• I think there's a typo; the triple product should be equal to $B(A \cdot C) - (C \cdot A)B$? Nov 14, 2021 at 10:37
• @Vtand : you are right. This expression can be written $B \alpha - C \beta$ (without the dot product, just multiplication of a vector by a scalar, therefore is a vector, not a scalar. Nov 14, 2021 at 10:51
• @Unknown In this case Griffith's Dynamics is wrong : the expression given by Vtand is the only right one... : if you examine the expression you have written, it gives always a multiple of $B$. Nov 14, 2021 at 10:55
• @JeanMarie Yep! You are really correct. I have rechecked the book. Now I found it.. Sorry for the mistake also.. :) Nov 14, 2021 at 15:45

$$A \times (B \times C) = B \underbrace{(A \cdot C)}_{\alpha \in \mathbb R} - C \underbrace{(A \cdot B)}_{\beta \in \mathbb R}=\alpha B - \beta C$$
is a vector belonging to plane defined by $$B$$ and $$C$$, therefore orthogonal to $$B \times C$$ which is very natural when you see its definition...
Its magnitude is $$\|\alpha B - \beta C \|$$.