Question

If $\overrightarrow{x}$ and $\overrightarrow{y}$ are two non-collinear 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

Answer: (C)

A right-angled triangle

Since $\overrightarrow{v},\overrightarrow{y}$ 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, $\Delta$ $ABC$ is right angled triangle.