Question

Assertion (A): If each of the angles $A,B,C$ is not a multiple of $\pi$, then the vectors \[ \vec r_1=(\sec^2A)\hat i+\hat j+\hat k, \] \[ \vec r_2=\hat i+(\sec^2B)\hat j+\hat k, \] \[ \vec r_3=\hat i+\hat j+(\sec^2C)\hat k \] are coplanar. Reason (R): The three vectors \[ \vec a=a_1\hat i+a_2\hat j+a_3\hat k, \] \[ \vec b=b_1\hat i+b_2\hat j+b_3\hat k, \] \[ \vec c=c_1\hat i+c_2\hat j+c_3\hat k \] are coplanar if and only if \[ \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix}=0. \] Which of the following is true?

• (A) is true, (R) is true and (R) is a correct explanation of (A)
• (A) is true, (R) is true but (R) is not a correct explanation of (A)
• (A) is true but (R) is false
• (A) is false but (R) is true

Hint:

Whenever a coplanarity question appears, immediately think of the determinant (scalar triple product) test.

Answer 1

The Correct Option is D

Step 1: Concept
Three vectors are coplanar if and only if their scalar triple product is zero.

Step 2: Meaning
The determinant criterion stated in the Reason is a standard and correct test for coplanarity.

Step 3: Analysis
For the given vectors, \[ \Delta= \begin{vmatrix} \sec^2A & 1 & 1\\ 1 & \sec^2B & 1\\ 1 & 1 & \sec^2C \end{vmatrix}. \] There is no general identity that makes this determinant zero for arbitrary angles $A,B,C$. Hence the vectors are not necessarily coplanar. Therefore the Assertion is false. However, the Reason correctly states the coplanarity criterion using determinants.

Step 4: Conclusion
Thus Assertion is false, while Reason is true.

Final Answer: (D)