Question
For a triangle $ABC$, let $\overrightarrow{p} = \overrightarrow{BC},\overrightarrow{q} = \overrightarrow{CA}$ and $\overrightarrow{r} = \overrightarrow{BA}$. If $|\overrightarrow{p}| = 2\sqrt{3},|\overrightarrow{q}| = 2$ and $cos\theta = \frac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\overrightarrow{P}$ and $\overrightarrow{q}$, then $|\overrightarrow{p} \times (\overrightarrow{q} - 3\overrightarrow{r})|^{2} + 3|\overrightarrow{r}|^{2}$ is equal to:
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Solution
$${\overrightarrow{p} + \overrightarrow{q} = \overrightarrow{r} }{cos(\pi - \theta) = \frac{|\overrightarrow{p}|^{2} + |\overrightarrow{q}|^{2} - |\overrightarrow{r}|^{2}}{2|\overrightarrow{p}\|\overrightarrow{q}|} }{\frac{- 1}{\sqrt{3}} = \frac{12 + 4 - |\overrightarrow{r}|^{2}}{2 \cdot 2\sqrt{3} \cdot 2} }{\|\overrightarrow{r}\|^{2} = 24 }{\therefore|\overrightarrow{p} \times (\overrightarrow{q} - 3\overrightarrow{r})|^{2} + 3|\overrightarrow{r}|^{2} }{= |\overrightarrow{p} \times (\overrightarrow{q} - 3\overrightarrow{p} - 3\overrightarrow{q})|^{2} + 72 }{= |\overrightarrow{p} \times ( - 3\overrightarrow{p} - 2\overrightarrow{q})|^{2} + 72 }{= | - 2\overrightarrow{p} \times \overrightarrow{q}|^{2} + 72 }{= 4|\overrightarrow{p}|^{2}|\overrightarrow{q}|^{2} \times \sin^{2}\theta + 72 }{= 4 \cdot 12.4 \cdot \frac{2}{3} + 72 }{= 200}$$
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