Vectors · Cross product magnitude

For a triangle

ABC

, let

\vec p = \overrightarrow{BC}

\vec q = \overrightarrow{CA}

and

\vec r = \overrightarrow{BA}

. If

\lvert\vec p\rvert = 2\sqrt3

\lvert\vec q\rvert = 2

and

\cos\theta = \dfrac{1}{\sqrt3}

, where

\theta

is the angle between

\vec p

and

\vec q

, then

\left\lvert\vec p \times \left(\vec q - 3\vec r\right)\right\rvert^2 + 3\lvert\vec r\rvert^2

is equal to:

We corrected a printing error in the original paper

Source printed “the angle between $\vec P$ and $\vec q$ ” (capital $P$ ; the vector is $\vec p$ ).

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