Question

If $\left({\sec }^{2}A\right)\hat{i}+\hat{j}+\hat{k},\hat{i}+\left({\sec }^{2}B\right)\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+\left({\sec }^{2}C\right)\hat{k}$ are coplanar, then ${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C$ is

• A. $1$
• B. $2$
• C. $0$
• D. None of these

Answer 1

The correct option is D None of these

For given vectors to be coplaner,

$\left|\begin{array}{ccc}{\sec }^{2}A& 1& 1\\ 1& {\sec }^{2}B& 1\\ 1& 1& {\sec }^{2}C\end{array}\right|=0$

$C_1\to C_1-C_2,C_2\to C_2-C_3$

$\left|\begin{array}{ccc}{\tan }^{2}A& 0& 1\\ -{\tan }^{2}B& {\tan }^{2}B& 1\\ 0& -{\tan }^{2}C& {\sec }^{2}C\end{array}\right|=0$

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 coplaner.