Question

If the vectors $\left({\sec }^{2}A\right)\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\left({\sec }^{2}B\right)\hat{j}+\hat{k}$, $\hat{i}+\hat{j}+\left({\sec }^{2}C\right)\hat{k}$ are coplanar, then the value of ${\sec }^{2}A+{\sec }^{2}B+{\sec }^{2}C-{\sec }^{2}A{\sec }^{2}B{\sec }^{2}C$ is

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

Answer: (B)

$2$

If the vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are co-planar, then

$(\overrightarrow{a}\times \overrightarrow{b}).\overrightarrow{c}=0$

$\therefore$ $\left[\right({\sec }^{2}Ai+j+k)\times (i+{\sec }^{2}Bj+k\left)\right].(i+j+({\sec }^{2}C\left)k\right)=0$

$=\left[\right(1-{\sec }^{2}B)i+j(1-{\sec }^{2}A)+k({\sec }^{2}A{\sec }^{2}B-1\left)\right].(i+j+k({\sec }^{2}C\left)\right)=0$

$=1-{\sec }^{2}B+1-{\sec }^{2}A+{\sec }^{2}A{\sec }^{2}B{\sec }^C-{\sec }^{2}C=0$

$={\sec }^{2}A{\sec }^{2}B{\sec }^{2}C-{\sec }^{2}A-{\sec }^{2}B-{\sec }^{2}C+2=0$

$={\sec }^{2}A+{\sec }^{2}B+{\sec }^{2}C-{\sec }^{2}A{\sec }^{2}B{\sec }^{2}C=2$