Question

If $\overrightarrow{x}$ and $\overrightarrow{y}$ are two non-parallel vectors and $ABC$ is a triangle with side lengths $a,b$ and $c$ satisfying $(20a-15b)\overrightarrow{x}+(15b-12c)\overrightarrow{y}+(12c-20a)(\overrightarrow{x}\times \overrightarrow{y})=\overrightarrow{0}$, then triangle $ABC$ is

• A. an acute-angled triangle
• B. an obtuse-angled triangle
• C. a right-angled triangle
• D. an isosceles triangle

Answer 1

The correct option is C a right-angled triangle

Since, $\overrightarrow{x},\overrightarrow{y}$ and $\overrightarrow{x}\times \overrightarrow{y}$ are linearly independent, we have
$20a-15b=15b-12c=12c-20a=0$
$\Rightarrow \frac{a}{3}=\frac{b}{4}=\frac{c}{5}$
$\Rightarrow c^2=a^2+b^2$
Hence, $△ABC$ is right angled.