Question

If the vectors $r_1={\sec }^{2}A,1,1;r_2=1,{\sec }^{2}B,1;$$r_3=1,1,{\sec }^{2}C$ are coplanar, then${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C$ is equal to

• A. 0
• B. 1
• C. 2
• D. not defined

Answer 1

The correct option is D not defined

For given vectors to be coplaner,
$\left|\begin{array}{ccc}{\sec }^{2}A& 1& 1\\ 1& {\sec }^{2}B& 1\\ 1& 1& {\sec }^{2}C\end{array}\right|=0$
$C_1\to C_1-C_2,C_2\to C_2\to C_3$
$\left|\begin{array}{ccc}{\tan }^{2}A& 0& 1\\ -{\tan }^{2}B& {\tan }^{2}B& 1\\ 0& -{\tan }^{2}C& {\sec }^{2}C\end{array}\right|=0$
Expanding along $R_1$
${\tan }^{2}A({\tan }^{2}B.{\sec }^{2}C+{\tan }^{2}C)+{\tan }^{2}B.{\tan }^{2}C=0$
$\Rightarrow {\tan }^{2}A.{\tan }^{2}B(1+{\tan }^{2}C)+{\tan }^{2}A{\tan }^{2}C+{\tan }^{2}B.{\tan }^{2}C=0$
$\Rightarrow {\tan }^{2}A.{\tan }^{2}B+{\tan }^{2}A{\tan }^{2}C+{\tan }^{2}B.{\tan }^{2}C=-{\tan }^{2}A.{\tan }^{2}B.{\tan }^{2}C$
Now divide both sides by, ${\tan }^{2}A.{\tan }^{2}B.{\tan }^{2}C$
$\Rightarrow {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C=-1$, which is not possible
Hence ${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C$ is undefined.
In another word given vectors can't be coplanar.