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: Understand assertion and reason.
The Assertion (A) claims three special vectors are always coplanar. The Reason (R) states the standard test: vectors are coplanar exactly when their determinant is zero. We judge each separately.
Step 2: Check the Reason.
It is a well-known true fact that three vectors are coplanar if and only if their scalar triple product, which equals that $3\times3$ determinant, is zero. So Reason is true.
Step 3: Build the determinant for the Assertion.
The vectors give \[ \Delta=\begin{vmatrix}\sec^2A&1&1\\1&\sec^2B&1\\1&1&\sec^2C\end{vmatrix}. \] Coplanar would need $\Delta=0.$
Step 4: Test with sample angles.
There is no identity forcing this to be zero for general $A,B,C.$ Try simple angles: with $A=B=C$ equal to some value not making cosine special, $\Delta$ comes out nonzero.
Step 5: Conclude about the Assertion.
Since $\Delta\ne0$ in general, the vectors are not always coplanar. So the Assertion is false.
Step 6: Pick the correct option.
Assertion false, Reason true. \[ \boxed{\text{A is false, R is true}} \]