For a triangle ABC, let p= BC,q= CA and r= BA.

Question

For a triangle $ABC$, let $\vec{p}=\overset{	o \leftarrow }{BC},\vec{q}=\overset{	o \leftarrow }{CA}$ and $\vec{r}=\overset{	o \leftarrow }{BA}$. If $|\vec{p}|=2\sqrt{3},|\vec{q}|=2$ and $\cos 	heta =\frac{1}{\sqrt{3}}$, where $	heta $ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p}	imes (\vec{q}-3\vec{r}){|}^{2}+3|\vec{r}{|}^{2}$ is equal to :

Accepted Answer

$\begin{gathered} \Rightarrow \vec{r}=\vec{q}+\vec{p} \ |\vec{r}|=\sqrt{{q}^{2}+{p}^{2}+2pq\cos 	heta } \ |r{|}^{2}=4+12+2	imes \sqrt{3}	imes 2	imes \frac{1}{\sqrt{3}} \ |r{|}^{2}=24 \ {|\vec{p}	imes (\vec{q}-3(\vec{q}-\vec{p}{)}^{2})|}^{2}+3|\vec{r}{|}^{2}=|-2\vec{p}	imes \vec{q}{|}^{2}+3|\vec{r}{|}^{2} \ =4	imes |\vec{p}||\vec{q}{|}^{2}{\sin }^{2}	heta +3	imes 24=4	imes 12	imes 4	imes (\frac{2}{3})+72=128+72=200 \end{gathered}$

Solutions · 1