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. $\frac{\left|\vec{c}\right|^{2}}{ 2} - \left|\vec{a}\right| = 12$
• B. $\frac{\left|\vec{c}\right|^{2}}{ 2} - \left|\vec{a}\right| = 30$
• 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

The Correct Option is D

$\vec{a}\cdot\vec{b}= 24$
$\vec{a}+\vec{b}+\vec{c} = 0$
$\vec{a} = -\vec{b}-\vec{c}$
$144 = 48+\left|\vec{c}\right|^{2} + 2\vec{b}\cdot\vec{c}$
$96 = \left|\vec{c}\right|^{2} + 2\left(24\right)$
$\left|\vec{c}\right|^{2} = 48$
$\left(A\right) \frac{\left|\vec{c}\right|^{2}}{2} -\left|\vec{a}\right| = \frac{48}{2}-12 = 12$
So option $\left(A\right)$ is correct
$\left(B\right) \frac{\left|\vec{c}\right|^{2}}{2}+\left|\vec{a}\right| = \frac{48}{2} + 12 = 24$
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)$
$= 2\sqrt{48\left(48\right)-\left(24\right)^{2}}$
$= 2\times24\times\sqrt{3} = 48\sqrt{3}$
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