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Mathematics Question on Determinants

In a triangle ABCABC if ∣1ab 1ca 1bc∣=0\begin{vmatrix}1&a&b\\\ 1&c&a\\\ 1&b&c\end{vmatrix} = 0, then sin⁡2A+sin⁡2B+sin⁡2C=\sin^2 A + \sin^2 B + \sin^2 C =

A

4−9\frac{4}{-9}

B

94\frac{9}{4}

C

333 \sqrt{3}

D

1

Answer

94\frac{9}{4}

Explanation

Solution

We have ∣1ab 1ca 1bc∣=0\begin{vmatrix}1&a&b\\\ 1&c&a\\\ 1&b&c\end{vmatrix} = 0,
⇒(c2−ba)−a(c−a)+b(b−c)=0\Rightarrow (c^2 - ba) - a (c - a) + b (b - c) = 0
⇒c2−ba−ac+a2+b2−bc=0\Rightarrow c^2 - ba - ac + a^2 + b^2 - bc = 0
⇒a2+b2+c2−ab−bc−ac=0\Rightarrow a^2 + b^2 + c^2 - ab - bc - ac = 0
⇒12(a−b)2+(b−c)2+(c−a)2=0\Rightarrow \frac{1}{2} \\{ (a - b)^2 + (b - c)^2 + (c - a)^2\\} = 0
which is possible only when
     a−b=0,b−c=0,c−a=0\ \ \ \ \ a - b=0, b-c=0, c-a=0
∴   a=b=c\therefore \ \ \ a = b = c
Then, ∠A=∠B=∠C=60∘\angle A = \angle B = \angle C= 60^{\circ}
sin⁡2A+sin⁡2B+sin⁡2C=3sin⁡260∘\sin^2 A + \sin^2 B + \sin^2C = 3\sin^2 60^{\circ}
=3(32)2=94= 3 \bigg(\frac{\sqrt{3}}{2} \bigg)^2 = \frac{9}{4}