If x and y are two non-collinear vectors and A B C is a triangle with side lengths a, b, c satisfying (20 a-15 b) x+(15 b-12 c) y+(12 c-20 a)(x × y)= vec0 then triangle A B C is

Question

If x and y are two non-collinear vectors and A B C is a triangle with side lengths a, b, c satisfying (20 a-15 b) x+(15 b-12 c) y+(12 c-20 a)(x × y)= vec0 then triangle A B C is (A) an acute-angled triangle (B) an obtuse-angled triangle (C) a right-angled triangle (D) an isosceles triangle

Tardigrade Admin · Accepted Answer

Since x and y are linearly independent, 20 a-15 b =15 b-12 c =12 c-20 a=0 ⇒ (a/3)=(b/4)=(c/5) ⇒ c2=a2+b2 ⇒ Δ A B C is right-angled.

Q. If $x$ and $y$ are two non-collinear vectors and $A BC$ is a triangle with side lengths $a, b, c$ satisfying $(20 a - 15 b) x + (15 b - 12 c) y + (12 c - 20 a) (x \times y) = 0$ then $△ A BC$ is

an acute-angled triangle

an obtuse-angled triangle

a right-angled triangle

an isosceles triangle

Since $x$ and $y$ are linearly independent,
$20 a - 15 b$
$= 15 b - 12 c$
$= 12 c - 20 a = 0$
$\Rightarrow \frac{a}{3} = \frac{b}{4} = \frac{c}{5}$
$\Rightarrow c^{2} = a^{2} + b^{2}$
$\Rightarrow Δ A BC$ is right-angled.

Q. If x and y are two non-collinear vectors and ABC is a triangle with side lengths a,b,c satisfying (20a−15b)x+(15b−12c)y+(12c−20a)(x×y)=0 then △ABC is