Question

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

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

Answer 1

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

$a.b=-72$

Given, $|a|=12,|b|=4\sqrt{3}|a|=12,|b|=4\sqrt{3}a+b+c=0\Rightarrow a=-(b+c)$
We have, $|a{|}^2=|b+c{|}^2$
$\Rightarrow |a{|}^2=|b{|}^2+|c{|}^2+2b.c\Rightarrow 144=48+|c{|}^2+48\Rightarrow |c{|}^2=48\Rightarrow |c|=4\sqrt{3}\text{Also,}|c{|}^2=|a{|}^2+|b{|}^2+2a.b\Rightarrow 48=144+48+2a.b\Rightarrow a.b=-72$
$\therefore$ Option (d) is correct.
Also, $a\times b=c\times a$
$\Rightarrow a\times b+c\times a=2a\times b\Rightarrow |a\times b+c\times a|=2a\times b\Rightarrow |a\times b+c\times a|=2|a\times b|=2\sqrt{|a{|}^2|b{|}^2-(a.b{)}^2}=2\sqrt{\left(144\right)\left(48\right)-(-72{)}^2}=2\left(12\right)\sqrt{48-36}=48\sqrt{3}$
$\therefore$ Option (c) is correct.
Also,$\frac{|c{|}^2}{2}-|a|=24-12=12$
$\therefore$ Option (a) is correct.
$\text{and}\frac{|c{|}^2}{2}+|a|=24+12=36$
$\therefore$ Option (b) is not correct.