Question

In a triangle ABC, if a, b, c are the sides opposite to angle
A,B, C respectively, then value of $ \begin{vmatrix}b\,cos\,C&a&c\,cos\,B\\ c\,cos\,A&b&a\,cos\,C\\ a\,cos\,B&c&b\,cos\,A\end{vmatrix}$ is

• A. 1
• B. -1
• C. 0
• D. $a \cos A+b \cos B+c \cos C$

Answer 1

Answer: (C)

$ \begin{vmatrix}b\,cos\,C&a&c\,cos\,B\\ c\,cos\,A&b&a\,cos\,C\\ a\,cos\,B&c&b\,cos\,A\end{vmatrix}$
Apply $C_{1} \rightarrow C_{1}+C_{3}$ and using projection rule,
$ D= \begin{vmatrix}b\,cos\,C&a&c\,cos\,B+b\,cos\,C\\ c\,cos\,A&b&a\,cos\,C+c\,cos\,A\\ a\,cos\,B&c&b\,cos\,A+a\,cos\,B\end{vmatrix}$
$=\begin{vmatrix}b\,cos\,C&a&a\\ c\,cos\,A&b&b\\ a\,cos\,B&c&c\end{vmatrix}=0$