Question

Let $PQR$ be a triangle Let $\vec{a}=\overrightarrow{QR},\vec{b}=\overrightarrow{RP}and\vec{c}=\overrightarrow{PQ}$. If $\left|\vec{a}\right|=12,\left|\vec{b}\right|=4\sqrt{3}$ and $\vec{b}.\vec{c}=24$, then which of the following is (are) true ?

• A. $\frac{{|\vec{c}|}^2}{2}-\left|\vec{a}\right|=12$
• B. $\frac{{|\vec{c}|}^2}{2}-\left|\vec{a}\right|=30$
• C. $\left|\vec{a}\times \vec{b}+\vec{c}\times \vec{a}\right|=48\sqrt{3}$
• D. $\vec{a}\cdot \vec{b}=72$

Answer 1

Answer: (A), (C), (D)

$\vec{a}\cdot \vec{b}=24$
$\vec{a}+\vec{b}+\vec{c}=0$
$\vec{a}=-\vec{b}-\vec{c}$
$144=48+{\left|\vec{c}\right|}^2+2\vec{b}\cdot \vec{c}$
$96={\left|\vec{c}\right|}^2+2\left(24\right)$
${\left|\vec{c}\right|}^2=48$
$\left(A\right)\frac{{|\vec{c}|}^2}{2}-\left|\vec{a}\right|=\frac{48}{2}-12=12$
So option $\left(A\right)$ is correct
$\left(B\right)\frac{{|\vec{c}|}^2}{2}+\left|\vec{a}\right|=\frac{48}{2}+12=24$
So option $\left(B\right)$ is not correct
$\left(C\right)|\vec{a}\times \vec{b}+\vec{c}\times \vec{a}|=|\vec{a}\times \vec{b}-\vec{a}\times \vec{c}|$
$=\left|\vec{a}\times \left(\vec{b}-\vec{c}\right)\right|$
$=\left|\left(-\vec{b}-\vec{c}\right)\times \left(\vec{b}-\vec{c}\right)\right|$
$=\left|0+\vec{b}\times \vec{c}-\vec{c}\times \vec{b}+0\right|$
$=2\left|\vec{b}\times \vec{c}\right|$
$=2\left(\sqrt{{\left|b\right|}^2{\left|c\right|}^2-{\left(\vec{b}\cdot \vec{c}\right)}^2}\right)$
$=2\sqrt{48\left(48\right)-{\left(24\right)}^2}$
$=2\times 24\times \sqrt{3}=48\sqrt{3}$
So option $\left(C\right)$ is correct
$\left(D\right)\vec{a}+\vec{b}+\vec{c}=0$
$\vec{a}+\vec{b}=-\vec{c}$
$144+48+2\vec{a}\cdot \vec{b}=48$
$\vec{a}\cdot \vec{b}=-72$
So option $\left(D\right)$ is correct